If you choose a representation $W$ of $G_1$ which is isomorphic to $V_1$, then you can construct $V_2$ as $\mathrm{Hom}_{G_1}(W, V)$. But I don't think there can be a fully choice-free construction of $(V_1, V_2)$.
Here is a hazy argument; if you give me a rigorous definition of choice-free then I may able to do better. Suppose that I find $V_1$, $V_2$ and an isomorphism $V_1 \boxtimes V_2 \to V$. And suppose that you likewise find $V'_1$, $V'_2$ and $V'_1 \boxtimes V'_2 \to V$. If your construction is canonical, you should be able to give canonical isomorphisms $a_1: V_1 \to V'_1$ and $a_2: V_2 \to V'_2$, making the obvious diagram commute.
But the obvious diagram also commutes when $(a_1, a_2)$ is replaced by $(-a_1, -a_2)$. I don't see how you can possibly single out which of $a_1$ and $- a_1$ is better.