It depends on the size of the planet. A key concern is whether the body is in hydrostatic equilibrium. An object in hydrostatic equilibrium is approximately spherical, although it may become oblate due to rapid rotation. The question, then, is whether the division of the planet places it below the critical size required for hydrostatic equilibrium.
It turns out that there's no straightforward formula for this, but as a rule of thumb, a body with dimensions of 1500 km or more will become rounded, and a body with dimensions of 400 km or less will not (although this is composition-dependent!). In between the two is a transition zone. It seems likely that if the remnant of the planet is less than 400 km across, it will be stable; if not, it's unlikely to keep its shape.
I'll go out on a limb and say that if the body was already stable, it will remain so; if it was already in hydrostatic equilibrium, on the other hand, it will probably remain in hydrostatic equilibrium and be pulled into a sphere. The reason for this is that the precise threshold is currently not well known. I cited 400 km as a lower limit, but some rocky bodies may instead become spherical at 600 km while icy bodies become spherical at 400 km. Some authors propose even lower thresholds like 200 km. Essentially, only objects in the transition region (say, 600 km to 1500 km) are likely to shift from hydrostatic equilibrium out of it, and that's a fairly narrow range.
Of course, now that you've specified that the planet was originally twice the size of Earth, it seems clear that the fragments will, indeed, also be stable planets, as they are easily massive enough to be rounded by hydrostatic equilibrium - and thus, as per the IAU's definition, they can be planets (assuming they clear their respective orbits, of course). They will certainly be planetary-mass objects.
Rotation
It's worth talking about the spin of the fragments before and after the split. Initially, the planet is spinning at some angular velocity $\omega_o$ about an axis. It has a moment of inertia about that axis $I_p=\frac{2}{5}MR^2$, where $M$ and $R$ are its mass and radius. Then the angular momentum is
$$L_o=I_p\omega_o=\frac{2}{5}MR^2\omega_o$$
Angular momentum should be conserved after the collision. Say the split happens along the planet's equator (this is a simple case, really, but it preserves axial symmetry). Then each fragment also happens to have moment of inertia $I_f=\frac{2}{5}mR^2$, where $m$ is the mass of the fragment. By symmetry, the fragments should have the same angular speed $\omega_f$ and total angular momentum
$$L_f=I_f\omega_f+I_f\omega_f=2I_f\omega_f=\frac{4}{5}mR^2\omega_f$$
However, $m=M/2$, so
$$L_f=\frac{2}{5}MR^2\omega_f$$
and we can see that as $L_f=L_o$, $\omega_f=\omega_o$; that is, the rotation does not change.