Capacitor with different charges on each plate
Suppose you have two conductors kept at voltages $V_1$ and $V_2$ and they have charges $Q_1$ and $Q_2$ respectively. Then, one has the relation $$ Q_1 = C_{11}\ V_1 + C_{12}\ V_2\quad \textrm{and}\quad Q_2 = C_{12}\ V_1 + C_{22} \ V_2 \ , $$ which defines a (symmetric) Capacitance matrix that is determined by the geometries of the two conductors. This generalizes the case that one uses for capacitors. See Purcell's Electricity and Magnetism in the Berkeley Series in Physics for a discussion.
Note added: The assumption in the above formulation is that the conductors have finite size. Then, when the charges $Q_1$ and $Q_2$ are finite, one assumes that the potential at spatial infinity defines the zero of the potential.
Systems of plates are not typically considered capacitors unless they are globally neutral. Nevertheless, capacitance is a geometric property that is to do with the system more than the actual voltages and charges you apply to it, so that your question still makes sense: the capacitance is the same as it would be with symmetric charges.
More specifically, the (mutual) capacitance of two conducting surfaces is defined as the charge that must be applied to one surface so that the potential in the other one will rise by one unit; by energy considerations it must be symmetric.
If the charges on both surfaces are antisymmetric (i.e. $+Q$ and $-Q$) then there will be a potential difference between the plates of $V=Q/C$. If they are asymmetric, a similar statement holds: if plate 1 has charge $Q_1$ and plate 2 has charge $Q_2$, then there will still be a potential difference between them of $V=(Q_1-Q_2)/2C$.
The problem with this, though, is that you're no longer seeing the whole picture, and you'll also have to deal with the self capacitance of the plates, which wasn't a problem before: if you put 100 C of charge on one plate and 99 C on the other, there will still be some potential difference between the plates, but they are also at a very high potential with respect to anything else you might consider, and you will have a host of other problems. This is why the situation is hardly ever considered.
In the general case, then, you have not one but two independent voltages to consider; that is, the mean and the difference, or the two voltages of the plates separately. To deal with this appropriately, you need to use a general capacitance matrix as Suresh describes in his answer.
Finally, you also need to worry about what other charged systems you must consider, and where they are. You can't just conjure a large charge on one plate without taking it from somewhere, and depending on where you're grounding there may also be some significant energy of interaction there.
Suresh's answer gives the correct general formalism.
(1) For the specific case of a coaxial cable, the electric field between the two conductors is determined by the charge $-Q$ on the inner conductor, which terminates on $+Q$ worth of charge on the outer conductor. (There can't be any field inside the inner conductor, so all the field generated by its charge extends outward.) The potential difference between the two conductors will just be $Q/C$, where $C$ is the capacitance of the coax.
The "excess" $19Q$ on the outer coax conductor produces a field extending outwards from the coax. If this field extends to infinity the potential drop will be infinite. Instead, one can terminate the field on $-19Q$ worth of new charge situated somewhere outside the coax. If that new charge resides on a new third conductor, that new topology is characterized by a new capacitance $C_{new}$, which will set the potential difference between the outer coax and the new third conductors: $19Q/C_{new}$.
(2) It's interesting to generate the capacitance matrix for this configuration. (See suresh's answer for the capacitance matrix equation formulation.) Because the coax cable length is infinite, one must consider the "specific" charges and capacitances, that is, charge and capacitance per unit length.
First consider the infinite coax in isolation. Evaluating the matrix elements (with the inner conductor denoted "1") gives: $$C_{11}=-C_{12}=-C_{21}=C_{22}=C=\frac{2 \pi \epsilon}{\ln\left(r_2/r_1 \right)}$$
This matrix is degenerate, forcing $Q_1=-Q_2$.
Now add a third, grounded cylindrical conductor at radius $r_3$, $r_3>r_2$. $C_{11}$, $C_{12}$, and $C_{21}$ are unchanged, but $C_{22}$ acquires a new term:
$$ C_{22}= 2 \pi \epsilon \left( \frac{1}{\ln\left(r_2/r_1 \right)} + \frac{1}{\ln\left(r_3/r_2 \right)} \right) = C + C_{new}$$
This addition removes the degeneracy, allowing arbitrary charges to be assigned to the two coax's conductors, with the resulting voltages measured with respect to the grounded conductor. As $r_3$ increases, $C_{new}$ decreases, reaching the degenerate case in the limit.
Inverting this matrix equation and solving for the potential difference, one finds: $$ V_1 - V_2 = \frac{Q_1}{C} $$ agreeing with the first analysis result.
In general, the potential difference in terms of the capacitance matrix elements and charges is: $$ V_1 - V_2 = \frac{(C_{22}+C_{21}) Q_1 - (C_{11} + C_{12}) Q_2 }{C_{11} C_{22} - C_{12} C_{21} } $$