SageMath package: crystalline_obstruction.
$A = \operatorname{Jac}(y^2 = 4x^5 - 36x^4 + 56x^3 - 76x^2 + 44x - 23)$
$\det(1 - t\,31^{-1}\operatorname{Frob} \mid H^2_{\mathrm{dR}}(A/\mathbf{Q}_p)) = (t-1)^2(31t^4 + 48t^3 + 43t^2 + 48t + 31)/31$
Thus, $\rho\bigl(A_{\mathbf{F}_p^{\mathrm{al}}}\bigr) = 2$.
Compute 2 eigenvectors
$\begin{aligned} v_1 \equiv{}& \left(356,\,37,\,831,\,0,\,295,\,31\right) \pmod{31^2} \\ v_2 \equiv{}& \left(4,\,957,\,3,\,1,\,0,\,0\right) \pmod{31^2}. \end{aligned}$
The last coordinate of the vectors above gives the projection to $H^2/F^1$. Therefore, $v_1 \notin F^1$ and the corresponding algebraic cycle cannot lift to $\mathbf{Q}_p$.
Thus, we improved $\operatorname{rank} \operatorname{NS}(A_{\mathbf{Q}^{\mathrm{al}}}) \leq 2$ to $\operatorname{rank} \operatorname{NS}(A_{\mathbf{Q}^{\mathrm{al}}}) \leq 1$, and therefore $\operatorname{End}(A_{\mathbf{Q}^{\mathrm{al}}}) = \mathbf{Z}$.
van Luijk's method would have succeeded in this example by using a second prime.