Hmm. You've asked a classical question about a quantum-mechanical system, which makes it a bit hard to answer precisely. But the answer to the spirit of your question is "yes".

It's more precise not to think about subatomic particles in bound systems as "meandering" so much as having non-zero probability of being found in different places. And the probability of any scattering interaction, including nuclear decay, will depend (amongst other things) on the overlap of the initial and final state wave functions. So yes, it matters where the particles "meander."

A great example of this is K-capture, or inverse beta decay. This only happens almost entirely with S-shell electrons, since P and higher angular-momentum electron wave functions go to zero at the origin. Only an S electron "spends enough time near the nucleus" to be captured with any significant probability.

The density of a nucleus is about 1/6 of the density of its individual nucleons. That means that even within a nucleus, it is still mostly empty space. Still, being bound within that nucleus does have some effects on the nucleons, including some weakening of quark confinement. Hence the radius of individual protons and neutrons within the nucleus is slightly larger than for free protons and neutrons. They swell because the bag pressure is reduced. Further, while the primary mechanism for nuclear binding is virtual pion exchange, it is also true to some very small extent that individual quarks are free to wander from nucleon to nucleon. At extreme pressures, i.e., inside a neutron star or in high energy nuclear collisions, the quarks become free-er and free-er to wander, until in the limit of a quark gluon plasma they can wander across the entire extended volume of the nuclear matter. At normal nuclear temperature and pressure, this still happens, but only to a very tiny, practically immeasurably small, degree.