Question: Let as assume that matrix $M$ is a positive semi-definite Hermitian matrix. Is it true that there exists a Hermitian matrix $A$ such that $A^H A=A A=M$?

Answer: 
By unitary base change you may assume $M$ to be diagonal. Taking square roots of the (non-negative real) diagonal entries gives another real diagonal (hence Hermitian) matrix.

By the way square roots of square matrices are not unique in general, so the title is not really a well formulated question; certainly some square roots might be non-Hermitian.
Let $M$ be a $n\times n$ Hermitian matrix, then there exists a unitary matrix $U$ such that $U^\dagger U=I$ and $M=UDU^\dagger$ where $D$ is the diagonal matrix containing the eigenvalues $\{\lambda\_i\}$ of $M$ with $1\leq i \leq n$. As $M$ is Hermitian and positive semi-definite, $\lambda\_i\geq 0$. The square root $\sqrt{D}$ of the diagonal matrix $D$ is found by taking the positive square root of the elements $\{\lambda\_i\}$, which leads to $\sqrt {D}^2=D$ .

Then there is always one positive semi-definite Hermitian $N$ which satisfies $N^2=M$ and $N=U\sqrt{D}U^\dagger$.
\begin{align}
N^2&=\left(U\sqrt{D}U^\dagger\right) \left(U\sqrt{D}U^\dagger\right)\\
&=U\sqrt{D}U^\dagger U\sqrt{D}U^\dagger \\
&=U\sqrt{D}\sqrt{D}U^\dagger\\
&=UDU^\dagger\\
&=M
\end{align}
$N$ is Hermitian as
\begin{align}
N^\dagger&=\left(U\sqrt{D}U^\dagger\right)^\dagger\\
&=\left(U^\dagger\right)^\dagger\left(\sqrt{D}\right)^\dagger\left(U\right)^\dagger\\
&=U\sqrt{D}U^\dagger\\
&=N
\end{align}
Since, $N$ is unitary similar to the diagonal matrix $\sqrt{D}$, the eigenvalues of $N$ are $\sqrt{\lambda\_i}\geq 0$.

Therefore, N is a positive semi-definite Hermitian square root of M.