Change of electron spin in Carbon after Hybridization?
Short answer: No it didn’t. But hybridisation also never happened in the way it is drawn.

The typical hybridisation picture from textbooks that you have shown us, is not ‘the truth’, it’s not ‘an approximation of the truth’ and not even ‘somewhere near to the truth’. It is no more and no less than a set of drawings trying to use one concept to explain an experiment.

The experiment in this case is the structure of methane, which is found to be tetrahedral, while the concept is that of orbitals as standing waves and solutions of the Schrödinger equation. The latter says that in a carbon atom floating around in sheer nothingness, there should be five standing waves produced by the electrons, that were given the name 1s, 2s, 2px, 2py and 2pz. But this is an entirely hypothetical construct from our understanding — not many people have been to outer space and nowhere have people observed and experimented with single, unbound carbon atoms.

Carbon is only found in compounds (on Earth and in the near space, that is). But once it’s in a compound, you need to write down a whole new Schrödinger equation, that not only includes the carbon but also all the other atoms in the molecule. If you solve that equation, you will again get a set of standing waves, but waves, that are spread out across the entire molecule. You need to perfom quite a few mathematical tricks to reduce those waves back to the orbitals you calculated for a single carbon atom — but by doing so, you have lost the predictive power of where the electrons are to go. There is no longer a way to say if any electron is spin $\alpha$ or spin $\beta$.

For methane, calculations that there are in fact four bonding orbitals at two (!) different energy levels. If hybridisation occurred, one would expect only one energy level, because all carbon orbitals would have been made equal. The orbitals generally represent a carbon ground-state orbital added to the corresponding hydrogen orbitals of the same phase (carbon’s p-orbitals oriented so that the stab in between two hydrogens). The lower energy state of course corresponds to carbon’s 2s plus the four hydrogens. (See pictures in this German lecture script, part A, section 4.8)

Take a different molecule, e. g. formaldehyde. Drawing hybridisation schemes, you would expect the oxygen to be something like $\mathrm{sp}^2$ and have a ‘rabbit ear’-like pair of free electrons. But solving the Schrödinger equation does not give you any free electrons. Instead you find a multitude of orbitals, none of which exclusively surrounds the oxygen — excepting the one that represents oxygen’s 1s in simplified diagrams. Again, reducing this to any type of structure that would tell you anything about the ‘shape’ of the free oxygen electrons means that you would lose predictive power. (Luckily in that case, though, all electrons are spin-paired, so they will remain spin-paired.)