One way to formulate your problem is via Stiefel bundles.
In your case there is the bundle $SU(n) \to S^{2n-1}$ given by taking the first row vector of a special unitary matrix. You are asking for a way to reverse the process, i.e. if you have a unit row vector you want to complete it to not just a Hermitian basis but one that has determinant one.
The idea is to consider complex mirror reflections $M_p$ in the complex hyperplane orthogonal to vectors $p \in S^{2n-1}$. Let $q \in S^{2n-1}$ and let $p \in H_q S^{2n-1}$, this is meant to indicate the hemi-sphere of $S^{2n-1}$ centred at the point $q$. Then the composite
$$M_q \circ M_p$$
should be essentially the hermitian matrix you are looking for. Likely it will only be giving you the first column vector the one you want (depending on how you think about matrices, i.e. perhaps you will need to take the transpose). That vector will be the vector twice as far from p as q is, i.e. take the great circle from p to q, and go twice as far. That's why there is the discontinuity at the point antipodal to p.
Does this sound reasonable? Hmm, on second thought, I'm running a bit on autopilot here. I am not certain if you get every vector in $S^{2n-1}$ via this construction. In the orthogonal group $O_n \to S^{n-1}$ this construction works fine. I'll see if this construction can be fixed for $SU(n)$. But I might need to sleep on it.