How does the spectral response of a photodiode change with bias voltage?
The answer is hardly at all.

The overall quantum efficiency might be slightly higher (a few per cent?) due to the carriers being swept out of the depletion region quicker and having less chance to recombine before contributing to terminal current. 

The cut-off wavelength may be shifted very slightly so that if your optical wavelength is just on the edge of cut-off you could see a stronger effect. 

But neither of these effects is really strong enough to be characterized on a datasheet. Your uncertainty about the optical coupling efficiency from whatever beam or fiber is delivering light to the detector will be greater than the change in quantum efficiency. And the variation in cut-off wavelength due to temperature changes will probably be greater than (or at least comparable to) the effect due to bias voltage. 


  a biased one produces a larger signal to the same amount of light intensity, right?


Not really. One carrier pair per absorbed photon is one carrier pair per absorbed photon, regardless of bias voltage. 

The main trade-off is modulation frequency response vs noise. A biased photodiode will have slightly faster response than an ubiased one, but will produce more dark current (and thus more noise).