Let $A$ and $B$ be Hermitian positive definite matrices. Then, the following matrix
\begin{equation*}
\begin{pmatrix}
A & X\\
X^* & B
\end{pmatrix}
\end{equation*}
is positive definite if and only if $X=A^{1/2}ZB^{1/2}$ for some $Z$ that satisfies $\|Z\| \le 1$ (thus, the range of $X$ is a subspace of the range of $A$, and the range of $X^*$ is a subspace of the range of $B$).
To this matrix, apply the Schur-complement's lemma to conclude that
\begin{equation*}
A \ge X^* B^{-1}X.
\end{equation*}
Using $X=R^{-1/2}$ and $B=A^{-1}$, we must therefore have
$R^{-1/2} = A^{1/2}ZA^{-1/2}$ for some contraction $Z$. If $R$ satisfies such an equality, the original claim will be true.