This should never really work in dimension at least two.

Suppose $R$ has two elements $x$ and $y$ that satisfy no relations beyond the obvious and are non-zero-divisors. Two generators of $m/m^2$ in any ring of dimension at least $2$ should do.

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 $R$ ($a\to x$, $b\to y$) and so is no elements are $x$-torsion or $y$-torsion while the other module has $x$-torsion and $y$-torsion.

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.