This should never really work in dimension at least two.
Let $x$ and $y$ be linearly independent elements of $m/m^2$.
Consider the modules $R^2/(ya-xb)$ and $R\oplus R/(x,y)$. They both have the same fitting ideals $I_0(M)=0$, $I_1(M)=(x,y)$, $I_2(M)=1$. These modules are nonisomorphic because the kernels of the map $M \otimes R/m^2 \to M \otimes R/m$ have different dimensions as vector spaces over $R/m$ because there is a different number of relations, $1$ in the first case and $2$ in the second.