Yes, it's just because the black hole is smaller. The only mass factor relevant to the trajectory of a particle in a spherically symmetric mass distribution is the total mass that is closer to the center of mass than the particle. Consequently, a photon emitted near the center of a 100 solar mass star will eventually (after many, many interactions with the dense hot medium of the star) work its way to the surface, because at any given radius in Schwarzschild coordinates, the mass "under" the photon is much less than the mass associated with a black hole with that radius.

Yes, it's just because the black hole is smaller. The only mass factor relevant to the trajectory of a particle in a spherically symmetric mass distribution is the total mass that is closer to the center of mass than the particle. For any given photon at any given radius in Schwarzschild coordinates, the mass "under" the photon is much less than the mass associated with a black hole with that radius. Consequently, a photon emitted near the center of a 100 solar mass star is able to propagate outwards (although an individual photon will be almost immediately absorbed in the dense hot medium). The associated energy is transported as heat from the core to the surface after many interactions, in which photons are scattered, absorbed, and emitted.