Right, they'll load each other, though you'll get close to no interaction if you set R2 >> R1.
(Assuming a ladder network with series R1, shunt C1, series C2, shunt R2.)
Note that Q < 1, so you can't expect a very narrow band, with respect to the center frequency (i.e. Fc/BW < 1). So the "independent poles" approximation isn't too far off, in general -- for the general case where those Fc's are very different. Of course with more error when they're closer together, and more still when the impedances are similar.
All of which you should find in the transfer function, hidden cryptically as it is in that expression. Let's see... it should be of overall form s / (s^2 + w_0^2), with the zero at the origin (s) saying gain rises from DC to some point, then a pole cancels it out (one of the denominator factors) and that's your flat passband, then another pole sets the HF roll-off. You say there's two zeroes; that must be related to the loading effect, I think? I'd have to think about it a bit to figure how that should be... As a hand-wave, though, consider the case for C2 >> C1: in this case, for most (mid and high) frequencies, C2 is effectively a short and R2 acts in parallel with C1, thus acting in parallel further with R1 -- you have a resistor divider of R1 into R2, with a lowpass response given by R1 || R2 (i.e. the Thevenin resistance at C1), and C1. This represents the case for Fc(HP) << Fc(LP), a fairly useless condition, and/or Z2 << Z1, also fairly useless (strong loading effect). I'm not sure exactly what the consequence of this all is -- for practical values -- but it looks like it'll have a pole-zero cancellation effect, where for marginal values of these ratios (i.e. LP and HP too close, or significant loading effect), you get an asymmetrical passband.
Tempted to run through the equation myself and see; it's not too terrible a problem to work out. Only a pair of quadratics after all, as you say!
Also to see the other case, HP into LP (series C1, shunt R1, series R2, shunt C2), and the difference between them if any.
As for practical filter specs -- -3dB is where |H| = sqrt(2)/2), take the magnitude, equate, and solve for s -- there will be four solutions, as a quadratic of quadratics (i.e., it's s^4, but no s^3 or s terms, so really a quadratic in s^2), because anything that works for positive frequencies also works for negative.
Oh, uh, this isn't even quite the complete picture, because it should be -3dB with respect to passband peak, not absolute -- so you should solve for insertion loss first (d|H|/ds = 0, find maximum), and then use -3dB of that to solve for the above -- yeah, it can get pretty messy, unfortunately. If you find one or both of the above assumptions are handy (wide freq or impedance ratios), then you can apply those and some simplification should drop out (for one, insertion loss should be close to 0dB under those assumptions).
Aside, I usually work these in Fourier (complex, s --> jw), and I don't actually remember offhand if magnitude is anything special in Laplace domain? Taking the magnitude of a complex rational polynomial of course is rather a pain. Maybe I should use s more often...
Tim