This is a problem related to Mechanics. There is a physical quantity called the action, that is given by:
$$S[\mathbf x]=\int\limits_{t_A}^{t_B}L(t, \mathbf x(t), \dot{\mathbf x}(t))\mathrm dt\tag{1}$$
where
$$\dot{\mathbf x}(t):=\frac{\mathrm d}{\mathrm dt}\mathbf x\tag{2}$$
Hamilton's principle requires $\mathbf x$ (the trajectory of a particle) to be an extremum for $S$, subject to the conditions
$$\mathbf x(t_A)=\mathbf x_0, \ \ \ \frac{\mathrm d}{\mathrm dt}\mathbf x\Bigg\vert_{t=t_A}=:\dot{\mathbf x}(t_A)=\mathbf v_0.\tag{3}$$
But I believe something important would be that for a given $t_A, \mathbf x_0$ and $\mathbf v_0$. The solution $\mathbf x(t)\vert_{(t_A, t_B)}$, restricted to the interval of interest, should be independent of the choice of $t_B$. Otherwise it would be like the particle was "seeing into the future" to adjust its trajectory in order to make $S$ an extremal. This would make no sense in Physics.
My concern rises from looking at the problem of minimal surface of revolution. In that problem, the solutions to the Euler-Lagrange equation do not account for all points in the plane, I made a graph:
The black lines are some solution to the equation, and they are confined to a limited region in the plane. I could envision "stitching together" several catenaries to reach point $Q$, but the problem is that such a solution is no longer an extremum, the actual extremum is given by the red dotted line.
I know the underlying problems are somewhat different because in Mechanics we don't fix the $t_B$ endpoint, but looking at the figure, if I fix the starting point $P$, and the derivative then I have singled out one of the catenaries and I will never be able to reach point $Q$ unless I do the stitching... Even then, the solution is not an extremum.
So, my question is, is there a way to guarantee that for $t_A<t_B<t_C$, and given the initial conditions $\mathbf x_0$ and $\mathbf v_0$, the solutions to
$$S[\mathbf x]=\int\limits_{t_A}^{t_B}L(t, \mathbf x(t), \dot{\mathbf x}(t))\mathrm dt,\tag{4}$$
$$\bar{S}[\bar{\mathbf x}]=\int\limits_{t_A}^{t_C}L(t, \bar{\mathbf x}(t), \dot{\bar{\mathbf x}}(t))\mathrm dt,\tag{5}$$
satisfy the condition
$$\bar{\mathbf x}(t)\vert_{(t_A, t_B)}=\mathbf x(t).\tag{6}$$
Is it somehting that is trivially satisfied when you use "initial conditions" instead of "boundary conditions"? or is it necessary to impose some restriction on $L$?
In contrast to what is written in OP's eq. (3), Hamilton's principle, aka. the principle of stationary action, applies to boundary value problems (BVP), not initial value problems (IVP). Boundary conditions (BC) are necessary in order to derive Euler-Lagrange (EL) equations.
BVPs are not violating causality by "seeing into the future", but merely conditional statements of the form "if the past and future are such and such, then the period in between is such and such".
Concerning IVP vs. BVP, see also this related Phys.SE post.
OP's last question seems answered by appropriate uniqueness theorems for IVPs.