This should never really work in dimension at least two.
Suppose $R$ has two elements
Let $x$ and $y$ that satisfy no relations beyond the obvious and are non-zero-divisors. Two generators of $m/m^2$ in any ring be linearly independent elements of dimension at least $2$ should do.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 first module can be embedded into kernels of the map $R$ ($a\to x$, M \otimes R/m^2 \to M \otimes R/m$ have different dimensions as vector spaces over $b\to y$) and so R/m$ because there is no elements are $x$-torsion or a different number of relations, $y$-torsion while 1$ in the other module has $x$-torsion first case and $y$-torsion.2$ in the second.