You're absolutely right that in principle each scattered photon undergoes a change in energy due to the recoil of the object it scatters off. In Rayleigh scattering, that change is very small, though, and can generally be neglected. In general, when the energy of the photon is much smaller than the rest energy of the massive particle, the photon undergoes only a small change in energy.
Here's a qualitative argument for why this is true. Let $p$ be the momentum of the incoming photon. The momentum transferred to the massive particle is at most of order $p$ (specifically, it's at most $2p$). For a nonrelativistic particle, the kinetic energy is $p^2/(2m)$, so the energy transferred is of this order. So the fractional energy lost by the photon is of order
$$
{\Delta E\over E}\sim{p^2/2m\over pc}={pc\over 2mc^2}={E\over 2mc^2},
$$
where $E =pc$ is the original photon energy. So when $E\ll mc^2$, the change in energy is small in comparison to the original energy.
One common way to express this more quantitatively is via the Compton scattering formula,
$$
\Delta\lambda={h\over mc}(1-\cos\theta),
$$
where $m$ is the mass of the massive particle and $\theta$ is the angle through which the photon scattered. This formula is usually used for much higher-energy photons scattering off charged particles, but it is derived just from energy and momentum conservation, so it applies to any situation in which a photon scatters off a massive particle initially at rest.
If I've done the arithmetic right, when $m$ is the mass of an atom or molecule, $\Delta\lambda$ comes out to be of order $10^{-16}$ m, which is certainly small enough to be ignored for visible light.
To answer your last question, even though atoms and molecules are neutral, they are made of charged parts. The photon can and does interact with those parts.
