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 $\mathrm{Pow}(\omega)$. $B$ can be embedded in $\mathrm{Pow}(\omega)$ because $\mathrm{Pow}(\omega)$ has an independent subset of size $\omega$. Even of size $2^\omega$.
Don Monk
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