It depends on the initial states and on the box size.

Let us first consider two free charged particles - electron and positron for simplicity.

If initially their relative motion kinetic energy $E_{rel.}$ is smaller than the negative potential energy of their interaction $-e^2/|\mathbf{r}_1 -\mathbf{r}_2|$ at the distance $|\mathbf{r}_1 -\mathbf{r}_2|$ where you switch the interaction on, then the particles get coupled and form a bound state. They will move together and in this sense they become localized one with respect to the other, but the center of mass will still be free (non localized).

In the opposite case the charges fly apart from each other.

The same may happen in a box. The minimal kinetic energy in a box is proportional to $1/L^2$ and the first perturbative correction to it due to Coulomb interaction $V_{00}\propto\int \sin(\pi x_1)\sin(\pi y_1)\sin(\pi z_1)\cdot \sin(\pi x_2)\sin(\pi y_2)\sin(\pi z_2)\frac{d^3r_1d^3r_2}{|\mathbf{r}_1 -\mathbf{r}_2|}$ is proportional to $-1/L$. So for a large box the kinetic energy of their relative motion can be smaller than the negative potential energy of their attraction and they can get coupled relative to each other. The center of mass, however, will be moving "freely" in the box.

In the opposite case of small box the interaction will not be able to create a neutral system moving as **a whole** in this box. Similarly, if you box is large, but the initial state has high relative kinetic energy ($\propto n^2/L^2$), then no bound state creation is possible, although the total wave function will be distorted with respect to the initial $\sin(\pi n x_1)\sin(\pi n y_1)\sin(\pi n z_1)$ and $\sin(\pi n x_2)\sin(\pi n y_2)\sin(\pi n z_2)$. However, if you include radiative losses due to their mutual scattering (but not due to collisions with walls), the extra kinetic energy of their relative motion will be progressively lost until the charges get coupled and descend into the ground state of their relative motion.