The answer is negative. Let $A$ be the completion of the denumerable atomless BA $B$. Then $A$ is complete and atomless. $A$ can be isomorphically embedded in $\mathrm{Pow}(\omega)$. In fact, $B$ can be isomorphically embedded in $\mathrm{Pow}(\omega)$, and by Sikorski's extension theorem, this embedding can be extended to an embedding of $A$ into $\mathrm{Pow}(\omega)$.