Under the assumption that your graph is 2-connected, I think you can just proceed greedily.  In this case, we know that the boundary of each face of $G$ is in fact a cycle.  So, in a sense that can be made precise, there is a cycle that is 'closest' to $t$.  That is, just take the symmetric difference of all faces that are inside $t$ or incident with $t$.  If we are trying to pack vertex disjoint cycles that contain $t$, we might as well include this 'closest' cycle.  Now just recurse.