Each of the following implies that (the true) $\omega_1$ is inaccessible in $L$, and hence that there are only countably many constructible reals:
- The proper forcing axiom
- There is a Ramsey cardinal
- $0^#$ exists
- All projective sets are Lebesgue measurable
- All $\Sigma^1_3$-sets are Lebesgue measurable
(EDIT: These are just some of the well-known examples that came to my mind. This list is neither exhaustive nor canonical.)
The mere existence of a nonconstructible set, or even a nonconstructible real, does not imply that $\omega_1^L$ is countable. There are many forcing notions in $L$ which do not collapse $\omega_1$: adding one or many Cohen reals, destroying Souslin trees, etc. Each such forcing (over L) results in a model where $\omega_1=\omega_1^L$.
In fact, "Martin's axiom plus continuum is arbitrarily large" is consistent with $\omega_1^L=\omega_1$. (And But also with $\omega_1^L<\omega_1$.)
ADDED: Preserving $\aleph_1$ of the ground model (which may or may not be the constructible universe $L$) is a key component in many independence proofs concerned with the theory of the reals. The "countable chain condition", which is enjoyed by all the forcings I mentioned above, is a property of forcing notions that guarantees preservation of $\aleph_1$; there are several other (weaker) properties which also suffice, most prominently (Baumgartner's) "Axiom A" and (Shelah's) "properness".