Is $R$ finitely generated? Let $A$ be a commutative ring with identity. Given two submodules $R,S$ of $A^n(n\in\Bbb N)$ and suppose $S$ is finitely generated, if there exists an isomorphism of $A$-modules $A^n/R\simeq A^n/S$, is $R$ finitely generated?
My example (below) was wrong, I misinterpreted what OP was asking for. I'll leave it up so others don't get confused as I did.

No: Let $A = \mathbb{C}[x_1, x_2, \ldots, x_n, \ldots]$ be the polynomial ring in infinitely many variables over $\mathbb{C}$, considered as a module over itself (i.e. the $n$ in your question is just $1$). Let $R$ be the submodule generated the even-indexed variables $x_{2i}$ and let $S = \{0\}$. Then $A/R \cong A/S \cong A$ and $S$ is finitely generated, but $R$ is not finitely generated.
My example (below) was wrong, I misinterpreted what OP was asking for. I'll leave it up so others don't get confused as I did.

No: Let $A = \mathbb{C}[x_1, x_2, \ldots, x_n, \ldots]$ be the polynomial ring in infinitely many variables over $\mathbb{C}$, considered as a module over itself (i.e. the $n$ in your question is just $1$). Let $R$ be the submodule generated the even-indexed variables $x_{2i}$ and let $S = \{0\}$. Then $A/R \cong A/S \cong A$ and $S$ is finitely generated, but $R$ is not finitely generated.