Is it possible to reconstruct the wavefunction of a molecule from a collection of spectra? Spectra of a molecule can be calculated if the wavefunction is known. Is it possible to do the opposite?
 A: In practice it would be very difficult to truly reconstruct "the" wave-function from the spectra. For one thing: Which wave function? Measured spectral emission/absorption are the result of transitions between states (ground and excited, e.g.) and these states all have their own wavefunctions. 
Furthermore, a typical measured spectrum is measured over a given frequency range and only gives information about a sum over energy eigenvalues (only for that particular frequency range) weighted by a matrix elements (which may impose selection rules) .
Some spectra (e.g., the dynamic form factor $S(\vec q,\omega)$) do give more direct information about the wavefunction in both momentum and energy, and therefore would be more useful to try to practically glean info about the wave function... people do try to do this sort of thing...
But anyways, here's one (EDIT: for another way, see below the line) theoretical way to determine the wave-function from the spectrum: 
1) Measure the spectra with very high precision over all frequencies and thereby determine all the eigenvalues of the Hamiltonian (the spectrum is typically proportional to $\delta(\hbar \omega +E_0-E_n)$ with some broadening); [EDIT: according to one of the other answers, measuring all the eigenvalues and their degeneracies may not be possible, but lets suspend our disbelief for a moment to finish this argument, then I will give another below]
2) Hope that the matrix element selection rules didn't keep you from accessing some of the eigenvalues in step 1;
3) Now that you know all the eigenvalues, you know the full Hamiltonian;
4) Given the full Hamiltonian, you can (in principle) solve the full Schrodinger equation for all the wavefunctions, thereby determining what you wanted to know.

EDIT: Here's another method that could, in principle, allow you to extract the wavefunction from the spectrum. 
In this case, I will assume that the spectrum of interest is the multiply-differential inelastic scattering cross-section, which is well-known to be proportional to the dynamic form-factor $S(\vec q,\omega)$. Therefore, I will assume, as above, perfect measurement of the spectrum and thus perfect knowledge of the entire dynamic form-factor. 
Now, the expression for the dynamic form-factor is given by:
$$
S(\vec q,\omega)=\sum_f \left\|{<f|\rho_{\vec q}}|0>\right\|^2\delta(E_0+\hbar\omega-E_f)\;,
$$
where $\rho_{\vec q}$ is the density operator in momentum space. Thus it is clear that by integrating $S(\vec q,\omega)$ over all $\omega$, followed by some algebraic manipulation, and then a Fourier transform back to position-space, we can obtain the ground-state density $\rho_0(\vec r)$.
Well then, given that we have the ground state density, we can then just appeal to the Hohenberg-Kohn Theorem which says that the potential is a unique functional (up to a constant) of the ground state density. But (ah ha!) the potential is the only non-universal function in the expression for the Hamiltonian of a molecule.
Thus, we have recovered the full Hamiltonian and thus the full Schrodinger equation. Which can, in principle, then be solved for any of the wavefunctions we require. 
Finally, I will note that from personal experience in condensed matter physics, people who build models always do like to try and interpret the spectra in terms of wave functions. And figuring out the parameters of wavefunctions based the spectra is done quite often... Indeed, this practice has always been the case (as @Anna v points out)... However, I don't think there is a well-known and practical general method to go straightforwardly from the spectrum to the wavefunctions.
A: No, you cannot hear the shape of the drum.
Even if you have a list of the eigenvalues of the Hamiltonian, you cannot reconstruct the Hamiltonian. For example, these two Hamiltonians have the same eigenvalues:

The spectral line is the same for both systems. Assuming that the transition is from energy 2 to energy 1, the first system goes from a state that is not degenerate to a state that is degenerate while the second system does the opposite.

Update to address comments
The question is about molecules. Enantiomers produce the same spectral lines. The wavefunctions might be related by a symmetry operation but they are not the same.
Having a list of the differences of eigenvalues is just that. Having the eigenvalues says nothing about the eigenstates. The first part of the answer points out that degeneracies might still be present. Those degeneracies have real consequences and cannot be dismissed (e.g. heat capacity).
The main reason you cannot reconstruct the wavefunction is because you do not know the relationship between the energy eigenstates and the position eigenstates. Having the list of eigenvalues tells you the list of $|n\rangle$ but you need $\langle x|n\rangle$.
A: It amuses me how often the cart is set in front of the horse in answers at this site. 
It is the data that drive theoretical formulations, not theoretical formulations the data.
It was from the spectral lines of the hydrogen atom that the Schrodinger equation was established and the whole construct of the theory of quantum mechanics took off.
The series describing the spectral lines of the hydrogen atom  had long been established . That the solutions of the Schrodinger equation could give the series started the whole identification of an operator Hamiltonian  whose solution gave probability distributions for the electron orbitals, instead of the orbits of the Bohr atom 
Spectral lines are finger prints for specific atoms, that is how we know the content of the sun , many stars and the interstellar medium ( from absorption lines). It is true that as atoms become more complex the solutions become harder , but if one starts with the Schrodinger equation and an approximate potential one can get the wave functions that will fit a specific spectral series for a specific atom. For molecular complexes and other set ups it will be harder.
