If you mean $|X(s)|\leq C$ for $0\leq s\leq t$ and any $0<t\leq 1$,
then the answer is yes; if you want $|X(s)|\leq C$ for $0\leq s\leq 1$
and any $0<t\leq 1$, then the answer is no in general.

Indeed, let $Y_j=Y_j(s)$ for $j=0,1$ be the solutions of the differential equation with $Y_0(0)=Y_1'(0)=I$ and $Y_0'(0)=Y_1(0)=0$, where $I$ stands for the identity matrix. Then the solution $X(\cdot,t)$ of the boundary value problem, for a given $0<t\leq1$, is
$$
X(s,t) = Y_0(s) X_0 + Y_1(s) Z(t),
$$
where the matrix $Z(t)$ is determined by the condition $X_1= X(t,t)= Y_0(t) X_0 + Y_1(t) Z(t)$. The condition on unique solvability of the boundary value problems is $\det Y_1(t)\neq0$ for all $0<t\leq1$, and then
$$
Z(t) = Y_1(t)^{-1}\left(X_1 - Y_0(t) X_0\right).
$$
Now, $Y_1(s) = s I + O(s^2)$ as $s\to+0$. Hence, $Y_1(s)^{-1} = s^{-1} I + O(s^0)$. It follows that $Z(t) = t^{-1} (X_1 - X_0) + O(t^0)$ and
$$
C = \sup_{0\leq s\leq t\leq 1}|X(s,t)| < \infty,
$$
while $X(1,t) = t^{-1} Y_1(1)\left(X_1-X_0\right) + O(t^0)$ which explodes as $t\to+0$ provided that $X_0\neq X_1$.

To estimate the constant $C\geq0$ in dependence on $A$ is a different matter, as one needs to find a way to express the (nonlinear) condition $\det Y_1(t)\neq0$ for $0<t\leq 1$ in terms of $A$.