Is every complete Boolean algebra isomorphic to the quotient of a powerset algebra? Is every complete Boolean algebra isomorphic to a quotient, as a Boolean algebra, of some powerset algebra $\wp(X)$?
It is not true for arbitrary Boolean algebras, see the comments, or see my MathSE question.

I am also aware of the Loomis-Sikorski Theorem: Every $\sigma$-complete Boolean algebra is isomorphic to the a quotient $\mathbf{F}/\mathcal{I}$, where $\mathbf{F}$ is a $σ$-field of sets and $\mathcal{I}\subseteq\mathbf{F}$ is a $\sigma$-ideal. So this can be viewed as asking whether there is an equivalent theorem for complete Boolean algebras.
 A: This variation of the question comes from the comments on the original question.
The question is whether all (complete) BAs are isomorphic in the category BA to a quotient of a powerset algebra.

The Sikorski extension theorem guarantees that every complete BA is a quotient of a power set algebra.
The Theorem. (Sikorski) Let $A$ be a subalgebra of a Boolean algebra $B$, and let $f:A\to C$ be a homomorphism from $A$ to a complete Boolean algebra $C$. Then $f$ can be extended to a homomorphism $\widehat{f}:B\to C$.
Application. Let $A=C$ be a complete Boolean algebra, and represent it as a subalgebra of a power set BA $B={\mathcal P}(X)$. (Use Stone duality or Birkhoff's subdirect representation theorem for this.) The Sikorski theorem guarantees that that the identity function ${\sf id}\colon A=C\to C$ can be extended to a homomorphism $\widehat{\sf{id}}: B={\mathcal P}(X)\to C$. Since $\widehat{\sf id}$ extends the identity function, it is surjective. \\\
This argument shows that any complete BA is in fact a retract of a power set BA. The converse is easily seen to be true, so the class of complete BA's is exactly the class of retracts of power set BA's. 

Also, from the comments on the original question:

* There are incomplete BA's that are quotients of power set BA's. E.g. ${\mathcal P}(\omega)/\textrm{fin}$.

* Infinite free BA's are not quotients of power set algebras. 
