This loop is rotating with angular speed $\omega$ in the counterclockwise direction in a magnetic field. I need to find the differential area element (for the purpose of finding flux). Is it going to be $d\vec{S}=(\omega t + \phi_o)d\rho dz \hat{\phi}$?
No it will be simply $d\rho \ dz\ \hat{\phi}$. The magnitude of $\vec{dS}$ will not change as $t$ changes and the direction will be along $\hat\phi$ at all points and times.
Is $\hat{\phi}$ the normal direction? If yes, then it already contains the angle of rotation, and the answer is $${\rm d}\vec{S} = \hat{\phi} \,{\rm d}\rho {\rm d}z$$
where $\rho$ is the distance from the $\hat{z}$ axis.
So at any point in time the normal vector should be $$\hat{\phi} = \underbrace{ \mathrm{rot}(\hat{z}, \phi_0 + \omega t)}_{3\times3\text{ matrix}} \; \hat{x} $$
