= Solution
The <canonical momenta> are $\pi_k=\dot\phi_k$, so
$$
Q=\int d^3x\,(\phi_2\pi_1-\phi_1\pi_2).
$$
Different species commute at equal times, so these products have no ordering ambiguity and $Q$ is <Hermitian>. Write $\int_{\mathbf p}=\int d^3p/(2\pi)^3$. Substituting the <mode expansion of a free field> and using the spatial <Fourier transform> identity gives
$$
\int d^3x\,\phi_2\pi_1
=-\frac i2\int_{\mathbf p}
(a_{2,\mathbf p}+a_{2,-\mathbf p}^\dagger)
(a_{1,-\mathbf p}-a_{1,\mathbf p}^\dagger).
$$
Subtract the expression with species interchanged. The two-annihilator terms cancel after $\mathbf p\mapsto-\mathbf p$, as do the two-creator terms. Commuting different species in the remaining terms yields
$$
\boxed{Q=-i\int_{\mathbf p}
(a_{2,\mathbf p}^\dagger a_{1,\mathbf p}-a_{1,\mathbf p}^\dagger a_{2,\mathbf p}).}
$$
There is no infinite <vacuum> contribution to the <Noether charge> to subtract: the contractions that would generate a species-diagonal constant are absent.
The bosonic <canonical commutation relations> give
$$
[Q,a_{1,\mathbf p}^\dagger]=-ia_{2,\mathbf p}^\dagger,
\qquad [Q,a_{2,\mathbf p}^\dagger]=ia_{1,\mathbf p}^\dagger.
$$
Define the <creation operators>
$$
c_{+,\mathbf p}^\dagger=\frac{a_{1,\mathbf p}^\dagger-ia_{2,\mathbf p}^\dagger}{\sqrt2},
\qquad
c_{-,\mathbf p}^\dagger=\frac{a_{1,\mathbf p}^\dagger+ia_{2,\mathbf p}^\dagger}{\sqrt2}.
$$
Their <commutators> with $Q$ are $[Q,c_\pm^\dagger]=\pm c_\pm^\dagger$. Since $Q|0\rangle=0$, the <one-particle states> have
$$
\boxed{Q\,c_{\pm,\mathbf p}^\dagger|0\rangle
=\pm c_{\pm,\mathbf p}^\dagger|0\rangle.}
$$
The fixed-<momentum> states are delta-function normalized; a square-integrable <wave packet> of either species is a normalizable state with the same <Noether charge> <eigenvalue>. In this basis $Q=\int_{\mathbf p}(c_+^\dagger c_+-c_-^\dagger c_-)$, so it measures a difference of occupation numbers. The equal-time identity $i[Q,\phi_1]=\phi_2$, $i[Q,\phi_2]=-\phi_1$ independently checks the sign against <1/a>.
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