As a general criterion, these kind of inequalities hold at the level of singular values (which implies a version of the inequality where one of the sides is conjugated by a unitary/partial isometry), but they tend to fail at the plain operator level.

Here is a counterexample for $p=3$, $q=3/2$. Take
$$
A=\begin{bmatrix}1&0\\0&3\end{bmatrix},\qquad\qquad B=\begin{bmatrix}1&1\\1&1\end{bmatrix}.
$$
Then
$$
\frac{A^3}3+\frac{2B^{3/2}}3=\frac13\,\Big(\begin{bmatrix}1&0\\0&27\end{bmatrix}
+\begin{bmatrix}2\sqrt2&2\sqrt2\\2\sqrt2&2\sqrt2\end{bmatrix}\Big) =\frac13\,\begin{bmatrix}1+2\sqrt2&2\sqrt2\\2\sqrt2&27+2\sqrt2\end{bmatrix}.
$$
We have
$$
AB+BA=\begin{bmatrix}2&21\\21&40\end{bmatrix},
$$
so
$$
(AB+BA)^2=\begin{bmatrix} 20&32\\32&52\end{bmatrix}
$$
and then Wolfram Alpha tells us that
$$
|AB+BA|=\begin{bmatrix} 20&32\\32&52\end{bmatrix}^{1/2}=\frac2{\sqrt5}\,\begin{bmatrix}3& 4\\ 4& 7
\end{bmatrix}.
$$

Then, using Wolfram Alpha again, we see that
$$
\frac{A^3}3+\frac{2B^{3/2}}3-\frac12\,|AB+BA|
$$
is not positive, as it has a negative eigenvalue.