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Will Sawin
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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.

Will Sawin
  • 148.4k
  • 9
  • 324
  • 563