How is the "charge on a capacitor" defined when two plates are unequally charged?

The "charge on a capacitor", in terms of circuit theory, is equal to the amount of charge that would flow down a wire connecting one plate to the other, if a wire were so connected, until the current stopped. The current will stop when there is no potential difference across the capacitor. This will happen when there is no electric field inside the capacitor. For a parallel plate capacitor this will happen when both plates carry the same charge (with the same sign).

For example, if intially the charges on the plates are $Q_1$ and $Q_2$, then an amount $(Q_1 - Q_2)/2$ will flow from plate $1$ to plate $2$, with the result that the charge then remaining on plate $1$ will be $$ Q_1 - \frac{(Q_1 - Q_2)}{2} = (Q_1 + Q_2)/2 $$ and the charge remaining on plate $2$ will be $$ Q_2 + \frac{(Q_1 - Q_2)}{2} = (Q_1 + Q_2)/2. $$ At this point both plates have the same charge with the same sign, so there is no electric field between them, and no potential difference between them. Hence the charge on the capacitor is equal to $$ \frac{Q_1 - Q_2}{2} . $$

The usual case is $Q_2 = -Q_1$ and then the final charge is zero, and the amount that flowed was $Q = (Q_1 - Q_2)/2$ which is then equal to $Q_1$.

Two plates having unequal charges do make up a capacitor and they are no different from the common sense of capacitors that we bear in our mind.

In your case, we would consider the charge on the inner surface. There is a reason for this. To understand this, let's first characterize a capacitor. So what comes up in your mind when you hear the word "capacitor"? Well, I think of some sort of charge separation which results in an electric field which has some energy associated to it. And this energy should be our characterising criteria, because that's what capacitors are used for, storing energy.

So now, in this case, which charges are responsible for the electric field and thus the electric field energy? Clearly, the field between the plates only depends on the charge on the inner surface of the plates. And thus while discussing the charge on the capacitor, it is the inner charge which matters. If you change the outer charges such that the inner charges don't change, then the field will also remain unchanged.