= Solution
Write $\int_{\mathbf p}=\int d^3p/(2\pi)^3$ and $d_{\mathbf p}=a_{\mathbf p}-b_{\mathbf p}$. The nonzero <canonical commutation relations> are
$$
[a_{\mathbf p},a_{\mathbf q}^\dagger]
=[b_{\mathbf p},b_{\mathbf q}^\dagger]
=(2\pi)^3\delta^{(3)}(\mathbf p-\mathbf q),
$$
with cross-species brackets zero. The <commutator product rule> gives
$$
\begin{aligned}
[Q,a_{\mathbf q}]&=\int_{\mathbf p}d_{\mathbf p}[d_{\mathbf p}^\dagger,a_{\mathbf q}]=-d_{\mathbf q},\\
[Q,b_{\mathbf q}]&=\int_{\mathbf p}d_{\mathbf p}[d_{\mathbf p}^\dagger,b_{\mathbf q}]=+d_{\mathbf q},\\
[Q,a_{\mathbf q}^\dagger]&=\int_{\mathbf p}[d_{\mathbf p},a_{\mathbf q}^\dagger]d_{\mathbf p}^\dagger=d_{\mathbf q}^\dagger,\\
[Q,b_{\mathbf q}^\dagger]&=\int_{\mathbf p}[d_{\mathbf p},b_{\mathbf q}^\dagger]d_{\mathbf p}^\dagger=-d_{\mathbf q}^\dagger.
\end{aligned}
$$
Linearity of the <commutator> now yields, for the real coefficient $s$,
$$
\boxed{\begin{aligned}
[Q,a_{\mathbf q}+sb_{\mathbf q}]&=-(1-s)(a_{\mathbf q}-b_{\mathbf q}),\\
[Q,a_{\mathbf q}^\dagger+sb_{\mathbf q}^\dagger]&=(1-s)(a_{\mathbf q}^\dagger-b_{\mathbf q}^\dagger).
\end{aligned}}
$$
Strictly, the printed $Q$ contains an infinite constant in continuum <Fock space>. Work first with finitely many modes, or replace it by its <normal ordering> $:Q:=\int_{\mathbf p}d_{\mathbf p}^\dagger d_{\mathbf p}$. Their difference is a constant, so every displayed <commutator> and every conjugation below is unchanged. The latter convention fixes the <vacuum state> phase without an infinite phase ambiguity.
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