Is an atom lighter if we cool down its nucleus quarks? I've just watched The Origin of Mass, in which most of our mass is explained to be coming from the very high kinetic energy of the quarks making up neutrons and protons, via the mass-energy equivalence relation of Einstein.
From there, I wondered if slowing down the quarks would reduce the overall atom mass? (by slowing down, I thought about cooling down).
At the same time, I'm confused about using the mass-energy equivalence here to explain this phenomenon. My understanding is that the Lorentz factor is very high at high speed, but the actual mass doesn't change.
 A: Nucleons and most nuclei are already in their ground states with respect the strong and electromagnetic interactions: there isn't a configuration with less energy.1 Nuclei and nucleon-like hadrons2 that are not in such a ground state can be identified by the fact that they decay. 
Now, those nuclei that decay by a pure gamma channel are less massive after the decay than before. So it there is a reasonable interpretation in which the answer to your question is "Yes", but you have to understand that there is no arbitrary ability to cool these systems: once they are in their quantum ground state there exist no states of the system that are low energy (less massive).
As Ben Crowell notes in the comments, the fact that a state that still has kinetic energy can be the lowest state it is possible to get arise from quantum mechanical consideration. In the simplest interpretation, we're trying to hold all the component parts of the system in a limited region of space which means that their momentum has to be uncertain, but that means that they have—on average—kinetic energy.

1 Free neutrons are emphatically not in a ground state with respect to the weak interaction as indicated by the fact that they decay to protons plus leptons.
2 I have the delta's in mind here, but there are even heavier states.
