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 $GL(n,\mathbb F_p)$.

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.