Let $M$ be an $R$-module ($R$ commutative with unity) and suppose that given $\mathfrak{p}\subset R$ a prime ideal we have $M_{\mathfrak{p}}\cong R_{\mathfrak{p}}^n$.
Is then true that we can find an element $f \in R$ such that $M_f \cong R^n[f^{-1}]$?
Yes...assuming that $M$ is finitely generated (if not see Mohan's helpful comment below).
Hint: The isomorphism $R^n_{\mathfrak{p}}\to M_{\mathfrak{p}}$ sends the basis vectors $e_i$ to elements of the form $\displaystyle \sum_j f_{i,j} m_j$ where $f_i\in R_{\mathfrak{p}}$. This isomorphism really only needs to have that the $f_{i,j}$ make sense, and this really only requires inverting finitely many denominators.