What defines a proton as "labile" in NMR?
For your first question you may want to look up "proton exchange NMR", as partly addressed in for instance this post:  

Proton attached to N or O in HNMR spectroscopy

$\ce{^1H/^2H}$ exchange is a solvent catalyzed process that results in rupture of the original O-H or N-H bond with formation of a new bond to previously H-bonded solvent $\ce{^2H/^1H}$. What keeps the exchange process from occuring is an activation energy barrier.  Available nonbonding electrons on O or N atoms in the solute or solvent which can interact with $\ce{^1H}$ or $\ce{^2H}$  affect the process, as does the pH (exchange can proceed through acid or based-catalyzed mechanisms). Catalysis of C-H bond rupture with formation of a new C-H bond has a larger activation energy barrier or may lead to chemically nonidentical products and reactants. 

As regards the second question, if the (protonated) solute is dilute and the solvent is deuterated, there isn't a significant concentration of solvent $\ce{^1H}$ to drive reversal to the protonated state. For instance, if your molecule is present in water at 1 mM concentration and contains one exchangeable proton, and if all of the protons exchange and the solvent water was originally 100% deuterated, then the final concentration of protons in the solvent is $\pu{\rho = ( 0.001/110) mole/mole = 9 \mu mole/mole (or 9 ppm)}$ and the solute will exhibit a similar final isotopic concentration ratio (ignoring isotope effects).