This can be solved with a smart use of Fubini's theorem:
$$
\begin{align*}
\int_a^{+ \infty} h(x)\, dx
&= \int_a^{+ \infty} \int_{x-a}^{x+a} f(t) \, dt dx \\
&= \int_a^{+ \infty} \int_0^{+ \infty} 1_{[x-a, x+a]} (t) f(t)\, dt \, dx \\
&= \int_a^{+ \infty} \int_0^{+ \infty} 1_{[t-a, t+a]} (x) f(t)\, dt \, dx \\
&= \int_0^{+ \infty} f(t) \int_a^{+ \infty} 1_{[t-a, t+a]} (x) dx dt.
\end{align*}
$$
Now, you can check that $\displaystyle\int_a^{+ \infty} 1_{[t-a, t+a]} (x)\, dx = \min \{t, 2a\}$. Hence,
$$
\begin{align*}
\int_a^{+ \infty} h(x)\, dx
&= \int_0^{+ \infty} f(t) \min \{t, 2a\}\, dt\\
&\leq 2a \int_0^{+ \infty} f(t)\, dt.
\end{align*}
$$
Edit : the method above is very useful and general. You can adapt it to see what happens of one puts $h(x) = \int_{x-a}^{x+a} g(x+t) f(t) dt$, where the function $g$ is, say, continuous. However, in your problem $g$ is equal to $1$, which is easier to deal with. Let $F$ be a primitive of $f$. Then :
$$
\begin{align*}
\int_a^X h(x) dx & = \int_a^X F(x+a)-F(x-a) \, dx \\
&= \int_{2a}^{X+a} F(x)\, dx - \int_0^{X-a} F(x)\, dx \\
&= \int_{X-a}^{X+a} F(x)\, dx - \int_0^{2a} F(x)\, dx.
\end{align*}
$$
Since $f$ is integrable, $F$ converges monotonically to $\int_0^{+ \infty} f(x)\,dx$, so that $\int_{X-a}^{X+a} F(x) dx$ converges to $2a \int_0^{+ \infty} f(x)\,dx$ (if you are not sure about that, you can write down the argument : for any $\varepsilon > 0$, there exists a $X> 0$ such that, for all $x > X$, we have $\int_0^{+ \infty} f(t)\,dt - \varepsilon \leq F(x) \leq \int_0^{+ \infty} f(t)dt$, and then...). Hence,
$$\int_a^{+ \infty} h(x)\, dx = \lim_{X \to + \infty} \int_a^X h(x)\, dx = 2a \int_0^{+ \infty} f(x)\,dx - \int_0^{2a} F(x)\, dx,$$
which is finite.