The short answer is "yes", and it's a special case of a much, much more general theorem on relatively free algebraic constructions.
In other language, you are asking whether the underlying functor from countably complete Boolean algebras to Boolean algebras has a left adjoint. The more general question is whether, given a homomorphism $\phi: S \to T$ between two monads on $Set$, the evident underlying functor
$$Set^\phi: Set^T \to Set^S$$
from the category of $T$-algebras to the category of $S$-algebras has a left adjoint. For this I'll direct your attention to this nLab article.
Of course, we have to know that the theory of countably complete Boolean algebras can be described as algebras of a monad on $Set$, but this too follows from general theory. I'll refer you to another nLab article for this; the article is not complete but it should give the idea. The upshot is that for any algebraic theory with only a small set of operations of each arity, there is a corresponding monad on $Set$ whose algebras are the models of the theory. The general constructions go back to work in the sixties, due to Lawvere, Linton, and others.
Edit: I'll remark that had you said "complete" instead of "countably complete", then the answer would have been no. In fact, the underlying functor from complete Boolean algebras to sets has no left adjoint; this is mentioned for instance in Categories for the Working Mathematician. But in your case, the theory is generated by a set of operations and equations, and all is well.