Let $R$ be a commutative ring and $M$ be a module with decreasing filtration $M=F_0M\supset F_1M\supset\dots$. Assume that $N$ is a $\mathbb Z$-graded $R$-module such that there is an isomorphism of graded $R$-modules
$$Gr_{\bullet}(M)\cong N,~~~~~~~~~~~~~~~~~~~~~~~~(1)$$
where $Gr_{\bullet}$ denotes the associated graded module. Does $(1)$ imply an isomorphism of $R$-modules $M\cong N$? I am actually interested in the case when $R$ is a field. Then, I think, the statement is valid but I would appreciate very much an idea of a proof and a more general context if possible.
There are easy counterexamples for non-fields. If $R=\mathbb{Z}$, for example, then the filtration $$ M = \mathbb{Z}/4\mathbb{Z} \supseteq \mathbb{Z}/2\mathbb{Z} \supseteq 0 $$ has associated graded $\mathop{gr} M = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, and $M \not \cong \mathop{gr} M$.
If $R$ is a field and if the filtration is Hausdorff (meaning that $\bigcap F_n M = 0$), then $\mathop{gr} M \cong M$ by dimension-counting. This is easy if $M$ is finite-dimensional, perhaps requires a bit more bookkeeping and care in the infinite-dimensional case.
If the filtration is not Hausdorff, then the claim can also fail. For example, the filtration $$ M \supseteq M \supseteq M \supseteq M \supseteq \cdots $$ (where all the inclusions are just equality) has associated graded $\mathop{gr} M = 0$. Similarly it can fail if you don't insist that $M = F_0 M$ (and I note that you do specify this): you need $\bigcup F_n M = M$ and $\bigcap F_n M = 0$.