Here is my original answer:
No; let $R=k[x_1,x_2,\ldots]$ for any field $k$ (a polynomial ring in infinitely many variables). Do you see a non-trivial ideal $J\subset R$ such that $R/J\cong R$? (There are a lot).
What I was aiming at was that, for any infinite set $S\subseteq\mathbb{N}$, choosing $J=(\{x_i\mid i\notin S\})$ gives a ring isomorphism $R/J\cong R$. The problem with my answer was that this is not the same as an isomorphism of $R$-modules. The definition of $R$-module is an abelian group that is acted on by $R$ (by scalar multiplication); the fact that $R/J\cong R$ as rings includes the fact that they are isomorphic as abelian groups (under addition), which is part of what is necessary for an isomorphism of $R$-modules, but as all the other (correct) answers point out, the essential problem lies in the scalar multiplication aspect: $J$ annihilates $R/J$ (i.e., scaling $R/J$ by any element of $J$ gives the zero map), while $R^n$ has trivial annihilator, so $J$ must be trivial.