Square root of doubly positive symmetric matrices

Robert Bryant already showed via an example that the answer is "no". To come up with lots of counterexamples, recall that (under some mild assumptions) if $A$ has maximal eigenvalue $\lambda_{\text{max}}$ and corresponding unit eigenvector $\mathbf{v}$ then $(A/\lambda_{\text{max}})^k \rightarrow \mathbf{v}\mathbf{v}^*$ as $k \rightarrow \infty$.

So if you pick any matrix that is (a) positive semidefinite with a negative entry, and (b) has a unique maximal eigenvalue with corresponding entrywise positive eigenvector, then repeatedly squaring $A$ will eventually give a counterexample to the original question.

No. If $$A = \begin{pmatrix}10&-1&5\\-1&10&5\\5&5&10\end{pmatrix},$$ then $A$ is positive definite but does not have all entries positive, while $$ A^2 = \begin{pmatrix}126&5&95\\5&126&95\\95&95&150\end{pmatrix} $$ is positive in both senses.