A much shorter proof, though one that doesn't explain as much to me. From $P^T F P = r^2 G$, we deduce $P^{-1} = r^{-2} G^{-1} P^T F$ and thus $P^{-1} \in \frac{1}{r^2 \det G} \mathrm{Mat}(\mathbb{Z})$. But also $P^{-1} = \frac{1}{\det P} \mathrm{Adj}(P)$ so $P^{-1} \in \frac{1}{r^n} \mathrm{Mat}(\mathbb{Z})$. Since $GCD(r^2 \det G, r^n) = r^2$, we have $P^{-1} \in \frac{1}{r^2} \mathrm{Mat}(\mathbb{Z})$, as desired.