Here's another approach; hopefully it adds something to the answers already given. Suppose there is a Borel set $E$, with $m(E)>0$, and a preassigned sequence $\{h_j\}$ such that $\chi_E(x+h_j)\not\to \chi_E(x)$ for a.e. $x\in E$. Then $f(x)=\chi_E(x)$ provides a counterexample. In the spirit of @Ben Derret's original answer, I'm trying to show that a much weaker conclusion fails to hold, even in the context of characteristic functions of Borel sets.
As for the existence of $E$, I think a fat Cantor set will work if the sequence $\{h_j\}$ is chosen carefully. More precisely, let $E=\bigcap_kE_k$, where $E_0=[0,1]$, and $E_{k+1}$ is obtained from $E_k$ by removing an open segment of length $\frac{1}{2}\cdot\frac{1}{3^{k+1}}$ from the center of each of the $2^{k}$ disjoint intervals whose union is $E_k$. It's not hard to show that $m(E)=\frac{1}{2}$.
Suppose $x\in E$, and let $I=[a,b]$ be the interval housing $x$ at the $k$th stage in the construction; i.e., $I$ is one of the $2^k$ disjoint intervals (of common length) whose union is $E_k$. The length of $I$ is then
$\frac{1}{2^k}\left(1-2^{-1}\sum_{n=1}^k\frac{2^{n-1}}{3^n}\right)=\frac{3^k+2^k}{6^k}<\left(\frac{5}{6}\right)^k\qquad(\text{assuming }k>1).$
So $x\in\left(b-\frac{i}{6^k},b-\frac{i-1}{6^k}\right]$ for some $i\leq 5^k$.
For $i=0,\dots,5^k$, define $p_{i,k}=\frac{i}{6^k}$, and let $\{h_j\}$ be an enumeration of the $p_{i,k}$ satisfying $h_j\to0$. (For instance, $p_{0,1},\dots,p_{5,1};p_{0,2},\dots,p_{25,2};\dots.$) If $x\in E$ and $I=[a,b]$ is as above, choose $p_{i,k}$ so that
$b
The complement $E_k^c$ of $E_k$ is the disjoint union of segments $S$, each with length at least $\frac{1}{2}\cdot\frac{1}{3^k}>\frac{1}{6^k}$. So $p_{i,k}+x$ lies in one of the $S\subset E_k^c$ (namely, the one that has $b$ for a left endpoint).
There is a $j$ for which $p_{i,k}=h_j$. Then $x+h_j\notin E_k\supset E$. This can be done for all $k$, so $x+h_j\notin E\,$ infinitely often, and, if I haven't made an error, $\chi_E(x+h_j)\not\to\chi_E(x)$ whenever $x\in E$.