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 $x_{i_k}\to x$).
Best Answer
I adapt the proof of a more general result: Theorem 4.12 in Chapter XI of K. Kuratowski and A. Mostowski’s book “Set Theory: with an introduction to descriptive set theory” (see page 408).
Claim. $f$ is essentially continuous iff there exists a null set $N$ such that $f|({\bf R}^n-N)$ is continuous.
Proof. ($\Rightarrow$) Let $U_1$, $U_2$, … be an open base of $\bf R$. By assumption, $$f^{-1}(U_n)=(O_n-A_n)\cup B_n,$$ where $O_n$ is open and $A_n$, $B_n$ are null sets. Set $N=\bigcup_nA_n\cup\bigcup_nB_n$, so that $N$ is a null set. We show that $g=f|({\bf R}^n-N)$ is continuous. Let $H\subset\bf R$ be open; we prove that $g^{-1}(H)=f^{-1}(H)-N$ is open in ${\bf R}^n-N$. Now $H=U_{k_1}\cup U_{k_2}\cup\cdots$, and so $$g^{-1}(H)=\bigcup_nf^{-1}(U_{k_n})-N=\bigcup_n\bigl((O_{k_n}-A_{k_n})\cup B_{k_n}\bigr)-N.$$ Since $A_{k_n}\cup B_{k_n}\subset N$, it follows that $$g^{-1}(H)=\bigcup_nO_{k_n}-N,$$ which completes the proof since $\bigcup_nO_{k_n}$ is open.
($\Leftarrow$) Let $N$ be a null set and suppose $g=f|({\bf R}^n-N)$ is continuous. Then, if $H\subset\bf R$ is open, the set $g^{-1}(H)=f^{-1}(H)-N$ is open in ${\bf R}^n-N$; that is, there exists open $O\subset{\bf R}^d$ such that $f^{-1}(H)-N=O-N$. Therefore, $$\begin{align*} f^{-1}(H) &= (f^{-1}(H)-N)\cup(f^{-1}(H)\cap N)\\ &=(O-N)\cup(f^{-1}(H)\cap N). \end{align*}$$ Since $N$ is a null set, $f^{-1}(H)\cap N$ is also a null set, and it follows that $f^{-1}(H)\triangle O$ is a null set. $\square$