The answer is no. Take countably many disjoint closed balls $B_i$ contained in the square $Q=[0,1]\times [0,1]$ and such that:

(i) Sum of areas of $B_i$ is less than 1

(ii) Sum of perimeters of $B_i$ is finite

(iii) $\bigcup B_i$ is dense in $Q$

Since the series $\sum \chi_{B_i}$ converges in BV norm, the set $E=Q\setminus \bigcup B_i$ has finite perimeter. It also has positive measure and empty interior. Any representative $F$ of the set $E$ also has empty interior and therefore $\partial F$ is not Lebesgue null.

By the way, any Lebesgue measurable set E has a representative F with the property

(*) $0<|F\cap B(x,r)|<|B(x,r)|$ for all $x\in\partial F$ and all $r>0$.

The proof is straightforward: add the points x for which $|E\cap B(x,r)|=|B(x,r)|$ for some r, and throw out all points x such that $|E\cap B(x,r)|=0$ for some $r$. (See Prop. 3.1 in "Minimal surfaces and functions of bounded variation" by E. Giusti.) By virtue of (*) the set $F$ has the smallest (w.r.t inclusion) topological boundary among all representatives of $E$, so if this representative doesn't help you, nothing does.