If $k$ is of characteristic zero or characteristic $p$ and $G$ is a pro-$p'$-group, then yes. In that case, everything in sight is semisimple by Maschke's theorem, the eigenvalues $a_i$ and $b_j$ give you the decomposition into irreducibles, and every pair of equal eigenvalues contributes one degree of freedom by Schur's lemma.

Otherwise no, even in the simplest case, because of lack of semiplicity. If $k=\overline{\mathbb{F}_p}$, $G=C_p$, $\phi_V(\phi) = \begin{pmatrix}1&1\\&1\end{pmatrix}$ and $\phi_W(\phi)=\begin{pmatrix}1&\\&1\end{pmatrix}$, then $a_1=a_2=b_1=b_2$ so that your conjecture would predict $\dim_k \operatorname{Hom}_G(V,W) = 4$, but it is only 2, because $V$ is not semi simple, but $W$ is, and therefore every morphism must factor through $V/\operatorname{rad}(V)$.

If $k$ is of characteristic zero or characteristic $p$ and $G$ is a pro-$p'$-group, then yes. In that case, everything in sight is semisimple by Maschke's theorem, the eigenvalues $a_i$ and $b_j$ give you the decomposition into irreducibles, and every pair of equal eigenvalues contributes one degree of freedom by Schur's lemma.

Otherwise no, even in the simplest case, because of lack of semiplicity. If $k=\overline{\mathbb{F}_p}$, $G=C_p$, $\rho_V(\phi) = \begin{pmatrix}1&1\\&1\end{pmatrix}$ and $\rho_W(\phi)=\begin{pmatrix}1&\\&1\end{pmatrix}$, then $a_1=a_2=b_1=b_2$ so that your conjecture would predict $\dim_k \operatorname{Hom}_G(V,W) = 4$, but it is only 2, because $V$ is not semi simple, but $W$ is, and therefore every morphism must factor through $V/\operatorname{rad}(V)$.