$\require{AMScd}$
A 1974 paper of R. Lagrange, Amalgamation and epimorphisms in $\mathfrak{m}$-complete Boolean algebras (Algebra Universalis 4 (1974), 277–279, DOI link), settled this affirmatively. In the cited paper, Lagrange shows that for any infinite cardinal $\mathfrak{m}$, the category of $\mathfrak{m}$-complete Boolean algebras with $\mathfrak{m}$-complete morphisms has the strong amalgamation property, which implies that epimorphisms are surjective. He remarks that the proof works just as well for complete Boolean algebras, and I'd also add that it can be adapted for plain old Boolean algebras. If I understand your meaning of an abstract $\sigma$-algebra correctly, this is the result you are after.
Let $\mathcal{C}$ be a concrete category so that we can meaningfully talk about epimorphisms being surjective.
$\mathcal{C}$ is said to have the strong amalgamation property if for every span $C \xleftarrow{f} A \xrightarrow{g} B$ of monomorphisms (aka amalgam), there exists an object $D \in \mathcal{C}$, and a commutative diagram of monomorphisms
$$
\begin{CD}
A @> g>> B\\
@VfVV @Vg'VV \\
C @>f'>> D,
\end{CD}
$$
such that $g'(B) \cap f'(C) = g'g(A) = f'f(A)$
Further restrict attention to a variety of algebraic structures, so that being a monomorphism is equivalent to mapping underlying sets injectively and every morphism canonically factors as a surjection followed by a monomorphism. Then the strong amalgamation property immediately implies that epis are surjective (see the corollary in Lagrange's paper).
I think this is a good — or at least thought-provoking — approach to addressing your question in light of your bounty comment that you are "looking for a canonical answer." For varieties, we see that the solution to the strong amalgamation problem always supplies a canonical non-unique extension: given a proper monomorphism $A \rightarrow B$, we get the strong amalgamation $D$ of $B \leftarrow A \rightarrow B$ together with distinct monomorphisms(!) $B \rightarrow D$ that agree on $A$.
Moreover, the solution to the strong amalgamation problem might be considered canonical in and of itself. Basically, Lagrange's method is a three-part construction: (1) Embed in the best coproduct available, which is in the ambient category of Boolean algebras (2) quotient that coproduct in order to force the desired intersection property of strong amalgamation (which has the awesome effect of restoring morphisms to our actual category) (3) Complete this quotiented coproduct, so that the whole embedding is now in-category. In other words, do the best you can with the coproduct you have... and then error-correct in the only sensible way. I guess that feels canonical.
On this last point, it might be interesting to compare this construction with solutions to the strong amalgamation problem in other varieties, in particular (finite) groups and Lie Algebras over fields.
If the Boolean algebra $B$ is complete, then I claim that there exists generalized limits. I am going to generalize this answer from limits indexed by $\omega$ to a limit of a net indexed by an arbitrary directed set.
Let $B$ be a complete Boolean algebra. Suppose that $D$ is a directed set. Then
let $I$ be the ideal on $B^{D}$ consisting of all $(b_{d})_{d\in D}$ where there exists some $e\in D$ where $b_{d}=0$ for all $d\geq e.$
Proposition: Suppose that $\varphi:B^{D}\rightarrow B$ is a Boolean algebra homomorphism. Then the following are equivalent:
$$\varphi((b_{d})_{d\in D})\leq\bigwedge_{d\in D}\bigvee_{e\geq d}b_{e}$$ for each tuple $(b_{d})_{d\in D}\in B^{D}$.
$$\bigvee_{d\in D}\bigwedge_{e\geq d}b_{e}\leq\varphi((b_{d})_{d\in D})$$ for each tuple $(b_{d})_{d\in D}\in B^{D}$.
$$\varphi(x)=0$$ whenever $x\in I$, and $\varphi((b)_{d\in D})=b$.
There is a Boolean algebra homomorphism $f:B^{D}/I\rightarrow B$ such that $f((b)_{d\in D}+I)=b$ for all $b\in B$ and where if $\pi_{I}:B^{D}\rightarrow B^{D}/I$ is the quotient mapping, then
$\varphi=f\pi_{I}.$
Proof outline: The equivalence between 1 and 2 follows by taking complementation. The equivalence between 3 and 4 follows from the first isomorphism theorem applied to Boolean algebras.
$(1\wedge 2)\rightarrow 3$. If $(b_{d})_{d\in D}\in I$, then there is some $d\in D$ where $b_{e}=0$ for $e\geq d$. Therefore,
$$\varphi((b_{d})_{d\in D})\leq\bigvee_{e\geq d}b_{e}=0.$$
Furthermore, we have $$\varphi((b)_{d\in D})\leq\bigwedge_{d\in D}\bigvee_{e\geq d}b=b\leq\bigvee_{d\in D}\bigwedge_{e\geq d}b\leq\varphi((b)_{d\in D}),$$ so
$\varphi((b)_{d\in D})=b.$
$(3\wedge 4)\rightarrow 1.$ Let $\varphi,\pi_{I},f$ be the mappings stated in the proposition where $\varphi=f\pi_{I}$.
Let $(b_{d})_{d\in D}\in B^{D}$, and suppose that $d_{0}\in D$. Let $b=\bigvee_{e\geq d_{0}}b_{e}$. Then
$\pi_{I}(b_{d})_{d\in D}\leq\pi_{I}((b)_{d\in D})$. Therefore,
$$\varphi(b_{d})_{d\in D}=f\pi_{I}(b_{d})_{d\in D}\leq f\pi_{I}((b)_{d\in D})
=\varphi((b)_{d\in D})=b=\bigvee_{e\geq d_{0}}b_{e}.$$
We conclude that
$$\varphi(b_{d})_{d\in D}\leq\bigwedge_{d_{0}\in D}\bigvee_{e\geq d_{0}}b_{e}.$$
Q.E.D
The mappings $f:B^{D}/I\rightarrow B$ such that $f((b)_{d\in D}+I)=b$ for all $b\in B$ can be obtained using Sikorski's extension theorem, so just let $\varphi=f\pi_{I}$ to obtain your generalized limits.
Best Answer
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 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.