Let's start by providing some useful background information that will help answer your question.
1,3-Butadiene is more stable than 2 separate ethylene molecules due to the extended conjugation possible in the diene. We can illustrate this by drawing simple resonance structures for the diene that can't be drawn for ethylene.
Huckel calculations show the same thing, 1,3-butadiene is more stable than 2 individual ethylenes. Here is the Huckel MO diagram for ethylene. The two electrons in ethylene have a net stabilization of 2$\beta$ compared to an electron in an isolated p-orbital. 4 electrons in 2 ethylenes would have an overall stabilization energy of 4$\beta$
Here is the Huckel diagram for 1,3-butadiene. The 4 electrons have a net stabilization of [(2 x 0.62) + (2 x 1.62)] = 4.48$\beta$.
Even though the HOMO in 1,3-butadiene is higher in energy then the HOMO for ethylene, overall 1,3-butadiene is more stable than 2 isolated ethylenes by 0.48$\beta$. This is because the MO below the HOMO is stabilized much more (0.62 $\beta$ compared to ethylene).
Now let's return to your question. Hyperconjugation is a reasonable way to explain why more highly substituted double bonds are more stable. Just like in 1,3-butadiene, when we delocalize electrons over a larger number of atoms, which is what hyperconjugation does, an overall stabilization of the molecule should result. And, as the hyperconjugated resonance structure shows, electron density will shift from the alkyl group to the double bond.
Even though the HOMO may be raised in energy, (at least some of) the other occupied MO's below the HOMO will be lowered in energy (just like in 1,3-butadiene) and overall a net stabilization of the molecule will result.