The conjectured inequality is **false**.

Using cyclicity of the trace, let's first write it in a slightly nicer form
\begin{equation*}
f(A,B) := \operatorname{tr}(X^*B^{(1-s)/2} A^sB^{(1-s)/2}X).
\end{equation*}

For joint-concavity to hold, we'd like to show
\begin{equation*}
f\left(\tfrac{A+B}{2}, \tfrac{U+V}{2} \right) \ge \tfrac12f(A,U)+\tfrac12f(B,V).\qquad (*)
\end{equation*}

But picking $A,B,U,V,X$ as follows, we have an immediate counterexample (I used $s=0.5$):
\begin{equation*}
A = \begin{pmatrix} 41 & 25\\ 25 & 26\end{pmatrix},\
B = \begin{pmatrix} 2 & 7\\ 7 & 25\end{pmatrix},\
U = \begin{pmatrix} 25 & 22\\ 22 & 20\end{pmatrix},\
V = \begin{pmatrix} 5 & 5\\ 5 & 10\end{pmatrix},\\
X = \begin{pmatrix} 1 & 0\\ 7 & -1\end{pmatrix}.
\end{equation*}

With this choice of variables, we obtain 5.964618e+03 < 6.529930e+03 in $(*)$ above.

## EDIT

In case you are interested or for others who may interested in concavity / convexity results similar in flavor to the one asked above, I'd like to mention the following nice paper: *Concavity of certain matrix trace and norm functions*, F. Hiai (2013).

In that paper, Hiai studies (among others), joint concavity and convexity of maps of the form
\begin{equation*}
(A,B) \mapsto \mbox{tr}(\Phi(A^p)^{1/2}\Psi(B^q)\Phi(A^p)^{1/2})^s,
\end{equation*}
for suitable choices of reals $p,q,s$, and positive linear maps $\Phi$ and $\Psi$ (e.g, with $\Phi(A) = X^*AX$, $\Psi=\mbox{Id}$, and $s=1$, we recover Lieb's concavity).