Let $g \in GL(n,\mathbb Q)$ normalize $GL(n,\mathbb Z)$. Consider the lattice $g(\mathbb Z^n) \subset \mathbb Q^n$; it is preserved by $GL(n,\mathbb Z)$. Replacing $g$ by $gz$ for some appropriate scalar matrix $z$, we may assume that $g(\mathbb Z^n) \subset \mathbb Z$, but that $g(\mathbb Z^n)\not\subset p \mathbb Z^n$ for any prime $p$.
Suppose now that $p$ divides the index $[\mathbb Z^n:g(\mathbb Z^n)]$. Then the image of $g(\mathbb Z^n)$ is a proper subspace of $\mathbb F_p^n$ (by the assumption that the $p$ divides the index) which is non-zero (by the assumption that $g(\mathbb Z^n)$ is not contained in $p\mathbb Z^n$). It is preserved by $GL(n,\mathbb F_p)$. [Added: As tomasz points out in a comment below, $GL(n,\mathbb Z)$ does not surject onto $GL(n,\mathbb F_p)$. However, its image does contain $SL(n,\mathbb F_p)$, so the argument below goes through, if we replace $GL(n,\mathbb F_p)$ by $SL(n,\mathbb F_p)$.]
But this is a contradiction, since $\mathbb F_p^n$ is an irreducible $GL(n,\mathbb F_p)$-representation. Consequently, no such $p$ exists, and so $g(\mathbb Z^n) = \mathbb Z^n$. Thus $g \in GL(n,\mathbb Z)$, and so we have shown that the normalizer of $GL(n,\mathbb Z)$ is equal to $Z(G) \cdot GL(n,\mathbb Z),$ as required.
This argument (assuming that it's correct!) extends at least to the case when $R$ is a PID.