Like any good quantum mechanical phenomenon, Raman scattering admits both a wave picture and a particle picture. You might not like either of them, though.
As far as the photon picture goes, it's simple: you have a scattering process that implements a unitary evolution $\hat U$ that takes a single photon $|1;\nu_0\rangle$ of energy $h\nu_0$ and the medium in its ground state $|g\rangle$ and produces a final state with a nonzero amplitude to have a lower-frequency photon $|1;\nu_0-\nu_v\rangle$ and the medium in its excited state $|e\rangle$, or in other words an evolution with a nonzero amplitude
$$\langle e|\langle 1;\nu_0-\nu_v| \ \hat U \ |g\rangle |1;\nu_0\rangle\neq 0,$$
or more pictorially
Now, if that big monolithic unitary $\hat U$ looks too mysterious, then you can apply the tools of leading-order perturbation theory to the problem (which, in the usual regime, can essentially be taken to be exact) and it will give you a nice expression in terms of virtual transitions. You don't like those? Too bad, because they're essential features of the description.
If you want to see this in terms of wave mechanics, and you're OK with confining yourself exclusively to the classical aspects of the EM emission, then the thing to look like is the medium itself: it is interacting with the field, and the field is inducing a coherent oscillation between the $|g\rangle$ and $|e\rangle$ states, which then amplitude-modulates the medium's oscillations, so that they're no longer monochromatic, and when you look at the oscillations of the medium's polarization, it develops AM sidebands which are precisely the Raman lines.
Now, the wave picture has a couple of problems. For one, you're pushing under the rug the description of the reaction of the medium to the light, even without the Raman-induced vibrational coherence, and if you want to correctly describe that polarization... you need virtual transitions, same as always. More importantly, though, you become blind to quantum properties of the emission, such as the fact that the Raman photons will normally be entangled with the pump photons, in experimentally-relevant ways that can only be explained in the photon picture. You can try to run away from quantum mechanics, if you want, but the only thing that you'll achieve is shut off doors that would otherwise stay open.