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 $\mathrm{Pow}(\omega)$.

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 $2^\omega$.

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 $Pow(\omega)$. In fact, $B$ can be isomorphically embedded in $Pow(\omega)$, and by Sikorski's extension theorem, this embedding can be extended to an embedding of $A$ into $Pow(\omega)$.

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 $\mathrm{Pow}(\omega)$.