Does the emission spectrum for a dye depend on the frequency of light used to excite the dye? Take a look at the emission spectrum for an arbitrary fluorescent dye: http://en.wikipedia.org/wiki/File:Fluorescein_spectra.jpg

Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$?  Consider that there is going to be a spread of eigenstates around a particular excited state (http://www.olympusmicro.com/primer/java/jablonski/jabintro/jablonskijavafigure1.jpg), corresponding to particular molecular vibration or rotational modes.  

The thought is that, if we excite a dye at a wavelength slightly red-shifted with respect to $\lambda_{max}$, or slightly blue-shifted with respect to $\lambda_{max}$, we'd start out at a different eigenstate clustered around the same excited state (say $S_1$).  I imagine there's a probability of decaying from each of these eigenstates in $S_1$, which gives us the spread for the emission spectrum, so if we bias the initial eigenstate, we'd achieve a different emission spectrum (presumably "statistically" red-shifted with respect to the excitation frequency).  Or is the fluorophore taking a rapid random walk through all accessible eigenstates belonging to a particular excited state?

I suppose we could also ask a similar question with respect to the photoelectric effect in terms of the energy provided to the excited electron (conditioned on having a threshold frequency to span a band gap).
Does the emission spectrum depend on exciting the dye at its peak absorbance wavelength $\lambda_{max}$? 


No, it is independent of that. The excitation $S_0 \rightarrow S_1$ is vertical, bond length are not changed in the excitation process. Compare the mass of the nuclei with that of the electrons and you know why. You'll just end up in a different vibronic level of the electronically excited state. 

However, before the emission takes place, this initial Franck_condon states relaxes to the vibrational ground state of $S_1$ with an equilibrated geometry and emission occurs from that (Kasha's rule).