No, there is no reason to suppose that $Q^{-1}BQ$ and $Q^{-1}CQ$ are the SAME diagonal matrix.

What you want to do FIRST, is prove two diagonal matrices commute:

$D_1D_2 = D_2D_1$ when $D_1,D_2$ are diagonal (PROVE THIS!).

Now if $Q^{-1}BQ = D_1$, then $QD_1Q^{-1} = Q(Q^{-1}BQ)Q^{-1} = (QQ^{-1})B(QQ^{-1}) = IBI = B$.

Similarly, if $Q^{-1}CQ = D_2$, then $QD_2Q^{-1} = C$.

Thus:

$BC = (QD_1Q^{-1})(QD_2Q^{-1}) = QD_1(Q^{-1}Q)D_2Q^{-1} = QD_1D_2Q^{-1} =\dots$ can you finish?