I will assume that being launched "straight up" means that the force exerted on the rocket is along a line between the rocket and the center of the planet. I will also assume that the rocket starts on the "equator" of the planet, i.e. it's position vector relative to the center of the planet is perpendicular to the planet's axis of rotation.
What this means is that there is no net torque acting on the rocket about the center of the planet, which means that its angular momentum about the center of the planet must be constant. Also, the rocket will continue to orbit about the same axis as it was before being launched.
Angular momentum in this case is given by
$$L=mr^2\omega$$
where $m$ is the mass of the rocket, $r$ is the distance the rocket is from the center of the planet, and $\omega$ is the rocket's angular velocity about the center of the planet.
The rocket starts at the surface of the planet a distance $R$ from the center of the planet with an angular velocity of $\omega_0$. Therefore, the initial angular momentum is
$$L_0=mR^2\omega_0$$
Once the rocket turns on, $r$ will be some increasing function of time $r(t)$. Since angular momentum is conserved, we have
$$L=L_0$$
$$mr^2\omega=mR^2\omega_0$$
Or, solving for $\omega$ as a function of time
$$\omega(t)=\left(\frac{R}{r(t)}\right)^2\omega_0$$
Therefore, as the rocket moves upwards, it's angular velocity will decrease. Therefore, it cannot stay at the same place above the planet. There would have to be some additional force tangential to the orbit in order to cause the angular velocity to remain constant, thus increasing the angular momentum as the rocket moves upwards.
