After reading your answer and thinking about it some more, I think I realize why I was initially confused about the use of class C biasing with push-pull. I think the drawing of the class C area of no conduction in the graphic included with the article is a little misleading. I understand what it's trying to convey, but I don't think it accurately reflects what you'd see if you were to observe the output of a single-ended class C amp on a scope, which is what I was thinking of when was looking at it.
The class C drawing is basically the inverse of the class AB drawing, and I can understand why: with class AB, the transistor is turned on for more than half the time, while with class C the transistor is turned on for less than half the time. However the drawing does not show the 'area under the curve,' as it were: it makes it seem as though there's just that bump sort of hanging there with nothing underneath it. Conceptually that make sense, but I think on a scope it would actually look like this (apologies for the blurriness):
(In a push-pull configuration, instead of the second half of the waveform being completely blacked out, you'd have another negative bump shaped just like the positive one.)
Basically, the transistor switches on as soon as the input reaches a certain level, at which point the output voltage shoots up suddenly in a straight line, rather than rising an falling gradually in sync with the input like it would do with class B biasing. And if I understand correctly, the resulting square waveform shape is what creates additional distortion products.
Mind you, what we're talking about here is the unfiltered output. I'm assuming that filtering the distortion products from the output also has the effect of smoothing out the waveform to restore its sinusoidal shape, so the filtered output signal would look much more like the input signal.
Of course, CB amplifiers almost never have any output filtering. 
-Bill
