No. Take a dense countable set $\{x_1,x_2,\dots\}$ in $\mathbb{R}^d$ and a sequence $(r_i)\subseteq\mathbb{R}^+$ such that $\sum_i r_i^{d-1}<\infty$. Then the function
$$f=1_{\bigcup_{i=1}^\infty B_{r_i}(x_i)}$$
is in $BV(\mathbb{R}^d)$ (since $|\bigcup B_{r_i}(x_i)|\le C\sum_i r_i^d$ and $f$ is the limit in $L^1$ of the functions $1_{\bigcup_{i=1}^k B_{r_i}(x_i)}$, whose gradients have total variation bounded by $C\sum_i r_i^{d-1}<\infty$).
Now, for any Lebesgue point $x_0$ of $f$ in the closed set $\{f=0\}$, no representative $g$ is continuous at $x_0$ (representative means a function which coincides a.e. with $f$). Indeed, $x_0$ lies in the closure of the open set $\bigcup B_{r_i}(x_i)$, so it belongs to the closure of $\{g=1\}$. On the other hand, since $x_0$ is a Lebesgue point for $f$, it must also belong to the closure of $\{g=0\}$. This shows that $g$ is not even a.e. continuous (since the set $\{f=0\}$ has positive measure).
Addendum.
The answer is still no even assuming $f$ continuous. Below I construct an example where the differentiability of $f$ fails on a Borel set of positive measure.
Choose a countable dense set $\{x_i\}$ in $B_1(0)$ and a sequence $r_i>0$ such that $\sum_i r_i^{d-1}<\infty$ and $\sum_i|B_{r_i}(x_i)|<|B_1(0)|$. In particular, $r_i\to 0$. Using Besicovitch covering theorem, up to a subsequence we can assume that the balls $B_{r_i}(x_i)$ have bounded overlapping (i.e. any point lies in at most $N$ such balls); in doing this, we could lose the density of $\{x_i\}$ but we still have $B_1(0)\subseteq\overline{\cup_i B_{r_i}(x_i)}$.
Let $S:=B_1(0)\setminus\cup_i\overline{B_{r_i}(x_i)}$. We remark that $|S|>0$ and that, for any $N\ge 1$, $S\subseteq\overline{\{x_i\mid i>N\}}$. It follows that we can find a sequence of positive radii $R_i\to 0$ such that
$S\subseteq\cup_{i\ge j}B_{R_i}(x_i)$ for all $j\ge 1$: by compactness, we can find $n_1>0$ such that
$$S\subseteq\overline{\{x_i\mid i>0\}}\subseteq\cup_{i=1}^{n_1}B_1(x_i),$$
then we use the above remark with $N=n_1$ and we find $n_2>n_1$ such that
$$S\subseteq\overline{\{x_i\mid i>n_1\}}\subseteq\cup_{i=n_1+1}^{n_2}B_{1/2}(x_i),$$
and so on.
Now the function $f(x):=\sum_i R_i(1-r_i^{-1}|x-x_i|)^+$ (a superposition of 'traffic cones' with heights $R_i$ placed on our balls $B_{r_i}(x_i)$) is continuous, as it is a uniform limit of continuous functions, thanks to the bounded overlapping. It also lies in $BV(\mathbb{R}^d)$ thanks to the assumption $\sum_i r_i^{d-1}<\infty$.
I claim that $f$ cannot be differentiable at $x$, for any $x\in S$. Indeed, as $x$ is a minimum point for $f$, we would have $\nabla f(x)=0$. But there is a sequence
$i_k\to\infty$ with $x\in B_{R_{i_k}}(x_{i_k})$, so $f(x_{i_k})\ge R_{i_k}\ge|x-x_{i_k}|$, which contradicts $\nabla f(x)=0$ (since $x_{i_k}\to x$).
There are strictly increasing $C^1$ functions that map sets of positive measure to sets of measure zero. Here is a construction:
Let $C\subset [0,1]$ be a Cantor set of positive measure. For a construction, see https://en.wikipedia.org/wiki/Smith-Volterra-Cantor_set. Let $g(x)=\operatorname{dist}(x,C)$. The function $g$ is clearly continuous and equal zero on $C$. In fact $g$ is a $1$-Lipschitz function. Let
$$
f(x)=\int_0^x g(t)\, dt.
$$
The function $f$ is $C^1$ and it is strictly increasing. Indeed, if $y>x$, then
$$
f(y)-f(x)=\int_x^y g(t)\, dx>0
$$
because the interval $[x,y]$ is not contained in the Cantor set $C$ and therefore it contains an interval where $g$ is positive.
On the other hand $f'=g=0$ on $C$ which has positive measure and $f(C)$ has measure zero since $m(f(C))=\int_C f'(t)\, dt=\int_C g(t)\, dt=0$.
As was pointed out by Mateusz Kwaśnicki in his comment, this construction gives the following result:
Theorem. Let $f$ be as above. Then there is a Riemann integrable function $h$ such that $h\circ f$ is not Riemann integrable.
Proof. The set $f(C)$ is homeomorphic to the Cantor set ($f$ is strictly increasing so it is a homeomorphism) and has measure zero as explained above. Let
$$
h(x)=\begin{cases}
1 & \text{if $x\in f(C)$}\\
0 & \text{if $x\not\in f(C)$.}
\end{cases}
$$
The function $h$ is Riemann integrable with the integral equal zero since it is bounded and continuous outside the set $f(C)$ of measure zero (because $\mathbb{R}\setminus f(C)$ is open and $h=0$ there). However,
$$
(h\circ f)(x)=\begin{cases}
1 & \text{if $x\in C$}\\
0 & \text{if $x\not\in C$.}
\end{cases}
$$
is not Riemann integrable since it is discontinuous on a set $C$ of positive measure. $\Box$
Best Answer
Let $F(x)=\int_0^x f(t)\, dt$. Then our assumption says that $$ G_r(x)\equiv \frac{F(x+r)+F(x-r)-2F(x)}{r^2} \to g(x) $$ as $r\to 0+$, uniformly on $E$. Thus also $G_r\to g$ in $\mathcal D'$ (= as distributions), by dominated convergence. However, in $\mathcal D'$, these difference quotients are just another way to obtain the distributional derivatives; this is well known in the simpler version for the first distributional derivative $u'$ of a $u\in\mathcal D'$, which we can obtain as $u'=\lim (\tau_h u-u)/h$, with the limit taken in $\mathcal D'$, but the argument is the same here: the operations are moved over to the test functions $\varphi$, and for these we have the required convergence to $\varphi''$ in the topology of $\mathcal D$.
We conclude that $F''=g$ in $\mathcal D'$, but $F'=f$, so we see that $f$ has an integrable distributional derivative and thus is absolutely continuous.