On the other hand, if the capacitor is finite, then the surface integrals over the plates are not equal.
No, but as you have shown it can be grossly overstated. The field outside a finite plate capacitor falls off as a power of distance >= 2 (details depend on the geometry), while the field inside it is constant. It can therefore safely be ignored for a first order estimate of the effect.
If you want to get fancy and claim that higher order corrections invalidate Zubrin's argument, you need to actually prove it. Also, don't forget to include other effects like plasma shielding.
Let A be the area of the capacitor, and dr the distance between the plates. Let c be the appropriate electrostatic constant for the coulomb force between a proton and the charge density on the plate. At a point a distance r from the capacitor, the field effect from the negative side is, ignoring curvature effects, about
cA/r^2
The repelling charge from the other plate will be about
cA/(r+dr)^2 = cA/(r^2 + 2rdr + dr^2) ~ cA(1/r)(1/(r+2dr))
So the difference between the coulomb forces, i.e. the net force, will be approximately
(cA/r)(1/r - 1/(r+2dr)) = (cA/r)((r+2dr - r)/(r^2 + 2rdr)) ~ cA(2dr/r^3)
So the net force drops off approximately as the third power of the distance, to a first order approximation. Integrating over the radius, we have that the potential goes as -1/r^2, with the approximation breaking down near r=0.
Actually inserting appropriate constants of integration would make this argument robust, but would also just reduce to the argument from potential at infinity. Either way it's clear that the effect can't just be ignored out of hand.
What really matters here is that with the force on the charge falling off as a power of distance, even if you integrate force * displacement from the screen out to infinity (which you shouldn't do in a plasma, because [2]), you get a finite contribution which can be made arbitrarily small relative to the work done inside the capacitor, where the force is constant, simply by increasing the size of the capacitor.
[1] http://student.ndhu.edu.tw/~d9914102/Teaching/EM/Paper/data/...
What's the dimension that you're proposing to increase of the capacitor? The total work done across the capacitor will be fixed regardless of distance across.
Since you insist: your derivation goes wrong right at the start by, as you say, "ignoring curvature effects", i.e. by considering radial distance only. By doing that, you are effectively imposing spherical symmetry; you are not doing parallel plates, you are doing concentric spherical shells. That makes the whole exercise pointless, since it's obvious from symmetry alone that such a device could never produce a net thrust: there is no preferred direction for the thrust to act along.
To answer your final question, just look at Eq. (19): make the plate radius (R) larger.
That answer should also be perfectly obvious from the limit case of infinite plates. A correct derivation for the general case must reproduce that result in that limit. Yours does the opposite; it gets worse the larger you make the capacitor. In the infinite limit, it is infinitely wrong.
You're also right that all that is moot from sheathing. But an ion still begins it's journey out at the bottom of a large potential well; one which is particularly steep because of the debye length, but still just as deep.