Corrected version
I believe it is always a square of an integer. We can assume that all our representations are over the algebraic closure $\overline{\mathbb Q}$ of $\mathbb Q$. If $\Gamma$ is the absolute Galois group, then clearly $\Gamma$ acts on the characters of $G$ (if you have a representation, then twist it by the action of $\Gamma$). It thus follows that the determinant squared is fixed by $\Gamma$ (since $\Gamma$ permutes the rows) and so is a rational number. But it is also an algebraic integer so it is an integer. Since you are squaring the determinant, you will get a perfect square.
