Let $f(x)=\frac{-1}{\log(x)+\log(1-x)}$ for $x\in(0,1)$ and $0$ for $x\not\in(0,1)$
Contour integration yields
$$
\begin{align}
\int_0^1\frac{-1}{\log(x)+\log(1-x)}e^{-2\pi ix\xi}\;\mathrm{d}x
&=\int_0^\infty\frac{-1}{\log(-is)+\log(1+is)}e^{-2\pi s\xi}(-i)\mathrm{d}s\\
&-\int_0^\infty\frac{-1}{\log(1-is)+\log(is)}e^{-2\pi i\xi-2\pi s\xi}(-i)\mathrm{d}s\\
&=\frac{1}{\xi}\int_0^\infty\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}e^{-2\pi s}(-i)\mathrm{d}s\\
&-\frac{e^{-2\pi i\xi}}{\xi}\int_0^\infty\frac{-1}{\log(1-is/\xi)+\log(is/\xi)}e^{-2\pi s}(-i)\mathrm{d}s\\
&\sim \frac{1-e^{-2\pi i\xi}}{2\pi i\xi\log|\xi|}\\
&=e^{-\pi i\xi}\frac{\sin(\pi \xi)}{\pi \xi\log|\xi|}
\end{align}
$$
as $|\xi|\to\infty$. $f$ is continuous and supported in $[0,1]$, so $f\in L^1$. $|\hat{f}(\xi)|\sim\left|\frac{\sin(\pi \xi)}{\pi \xi\log|\xi|}\right|$, so $\hat{f}\not\in L^1$.
Addition: If we replace $f$ by $f^\alpha$, the same proof yields $\widehat{f^\alpha}(\xi)\sim e^{-\pi i\xi}\frac{\sin(\pi \xi)}{\pi \xi(\log|\xi|)^\alpha}$ for $\alpha\ge0$. When $\alpha=0$, $f^\alpha$ is not continuous. Otherwise, $f^\alpha$ is a counterexample for $\alpha\in(0,1]$.
Asymptotics: First a couple of estimates. Note that
$$
\log(-it)+\log(1+it)=\frac{1}{2}\log(t^2+t^4)+i(\tan^{-1}(t)-\frac{\pi}{2}\operatorname{sgn}(t))
$$
Therefore,
$$
\begin{align}
|\log(-it)+\log(1+it)|
&\ge\left\{\begin{array}{}
\left|\tan^{-1}(t)-\frac{\pi}{2}\operatorname{sgn}(t)\right|\ge\frac{\pi}{4}&\text{when }|t|<1\\
\left|\frac{1}{2}\log(t^2+t^4)\right|\ge\frac{1}{2}\log(2)&\text{when }|t|\ge1
\end{array}\right.\\
&\ge\frac{1}{2}\log(2)\tag{1}
\end{align}
$$
for all $t$. The actual minimum is greater than $\frac{53}{62}$, but $\frac{1}{2}\log(2)$ is sufficient.
Furthermore, for $|t|<1$, we have
$$
\begin{align}
|\log(-it)+\log(1+it)|
&\ge|\log(|t|)|-\frac{1}{2}\log(1+t^2)\\
&\ge\log(1/|t|)-\frac{1}{2}\log(2)\\
&\ge\frac{1}{2}\log(1/|t|)\tag{2}
\end{align}
$$
for $|t|<\frac{1}{2}$.
Since $f(x)$ is the translate of an even function, $\hat{f}(\xi)$ is an even function times $e^{i\alpha\xi}$ for some $\alpha$. Therefore, an estimate for $\xi>0$ also works for $\xi<0$. To make notation neater, we will assume $\xi>0$.
Our goal is to estimate
$$
\int_0^\infty\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\;e^{-2\pi s}\;\mathrm{d}s\tag{3}
$$
asymptotically as $\xi\to\infty$. To this end, we will assume $\xi>1$ and split the domain of integration in $(3)$ into three parts: $I_1=[0,\xi^{-\epsilon})$, $I_2=[\xi^{-\epsilon},\xi^{\epsilon})$, and $I_3=[\xi^{\epsilon},\infty)$, where $\epsilon$ is chosen so that $\xi^{-\epsilon}=\epsilon$; that is, $\epsilon=\operatorname{W}(\log(\xi))/\log(\xi)\to0$ as $\xi\to\infty$. In fact, $\epsilon<\log(\log(\xi))/\log(\xi)$ for $\xi>e^e$.
$\underline{I_1=[0,\xi^{-\epsilon})}$: Since $|I_1|=\epsilon$ and $s<1$, estimate $(2)$ insures that for $\xi\ge2$,
$$
\int_{I_1}\left|\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\right|\;e^{-2\pi s}\;\mathrm{d}s\le\frac{2\epsilon}{\log(\xi)}\tag{4}
$$
$\underline{I_2=[\xi^{-\epsilon},\xi^\epsilon)}$: Let us break up the integral over $I_2$ into a few pieces.
$$
\begin{align}
&\int_{I_2}\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\;e^{-2\pi s}\;\mathrm{d}s\\
&=\int_{I_2}\frac{1}{\log(\xi)\left(1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}\right)}\;e^{-2\pi s}\;\mathrm{d}s\\
&=\int_{I_2}\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s+\int_{I_2}\left(\frac{1}{1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}-1\right)\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s\\
&=\frac{1}{2\pi\log(\xi)} - \int_{I_1+I_3}\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s + \int_{I_2}\left(\frac{\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}{1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}\right)\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s\tag{5}
\end{align}
$$
Because $|I_1|=\epsilon$, we get
$$
\int_{I_1}\left|\frac{e^{-2\pi s}}{\log(\xi)}\right|\mathrm{d}s\le\frac{\epsilon}{\log(\xi)}\tag{6}
$$
Because $I_3=[\xi^\epsilon,\infty)$, we get
$$
\begin{align}
\int_{I_3}\left|\frac{e^{-2\pi s}}{\log(\xi)}\right|\mathrm{d}s
&=\frac{e^{-2\pi\xi^\epsilon}}{2\pi\log(\xi)}\\
&=\frac{\epsilon e^{-2\pi/\epsilon+\log(1/\epsilon)}}{2\pi\log(\xi)}\\
&\le\frac{\epsilon}{2\pi\log(\xi)}\tag{7}
\end{align}
$$
For $s\in I_2$, $\left|\frac{\log(s)}{\log(\xi)}\right|\le\epsilon$ and $s/\xi\le1$. Furthermore, if $\xi>27$, then $\epsilon<\frac{1}{3}$; therefore, $\log(\xi)=\log(1/\epsilon)/\epsilon\ge\log(3)/\epsilon$.
$$
\begin{align}
\left|\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}\right|
&\le\left|\frac{\log(s)}{\log(\xi)}\right|+\left|\frac{\log(-i)+\log(1+is/\xi)}{\log(\xi)}\right|\\
&\le\epsilon+\left(\frac{\pi}{2}+\frac{1}{2}\log(2)\right)\frac{\epsilon}{\log(3)}\\
&\le\frac{11}{4}\epsilon\tag{8}
\end{align}
$$
Applying $(8)$ to the integral over $I_2$ in $(5)$, we get
$$
\begin{align}
\int_{I_2}\left|\frac{\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}{1-\frac{\log(-is)+\log(1+is/\xi)}{\log(\xi)}}\right|\;\frac{e^{-2\pi s}}{\log(\xi)}\;\mathrm{d}s
&\le\frac{\frac{11}{4}\epsilon}{1-\frac{11}{4}\epsilon}\frac{1}{2\pi\log(\xi)}\\
&\le\frac{33\epsilon}{2\pi\log(\xi)}\tag{9}
\end{align}
$$
$\underline{I_3=[\xi^\epsilon,\infty)}$: Using estimate $(1)$, we get
$$
\begin{align}
\int_{I_3}\left|\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\right|\;e^{-2\pi s}\;\mathrm{d}s
&\le\frac{2}{\log(2)}\frac{e^{-2\pi\xi^\epsilon}}{2\pi}\\
&=\frac{2}{\log(2)}\frac{\epsilon e^{-2\pi/\epsilon+\log(1/\epsilon)+\log(\log(\xi))}}{2\pi\log(\xi)}\\
&=\frac{2}{\log(2)}\frac{\epsilon e^{-2\pi/\epsilon+2\log(1/\epsilon)+\log(\log(1/\epsilon))}}{2\pi\log(\xi)}\\
&\le\frac{\epsilon}{\pi\log(2)\log(\xi)}\tag{10}
\end{align}
$$
Thus, combining equation $(5)$ with estimates $(4)$, $(6)$, $(7)$, $(9)$, $(10)$, we get that for $\xi>27$,
$$
\begin{align}
&\left|\int_0^\infty\frac{-1}{\log(-is/\xi)+\log(1+is/\xi)}\;e^{-2\pi s}\;\mathrm{d}s - \frac{1}{2\pi\log(\xi)}\right|\\
&\le(2+1+\frac{1}{2\pi}+\frac{33}{2\pi}+\frac{1}{\pi\log(2)})\frac{\epsilon}{\log(\xi)}\\
&\le\frac{6\epsilon}{\log(\xi)}\tag{11}
\end{align}
$$
Using estimate $(11)$ with its counterpart for $-\xi$, we get that
$$
\left|\int_0^1\frac{-1}{\log(x)+\log(1-x)}e^{-2\pi ix\xi}\;\mathrm{d}x-e^{-\pi i\xi}\frac{\sin(\pi \xi)}{\pi \xi\log|\xi|}\right|\le\frac{12\epsilon}{|\xi|\log|\xi|}
$$