= Solution
Define $F^\sharp(k)=\overline{F(\bar k)}$. The defocusing reduction of the <Lax pair> is $\mu(k)=\sigma_1\overline{\mu(\bar k)}\sigma_1$, so the <half-line NLS spectral functions> have the form
$$
s=\begin{pmatrix}a^\sharp&b\\b^\sharp&a\end{pmatrix},\qquad S=\begin{pmatrix}A^\sharp&B\\B^\sharp&A\end{pmatrix},\qquad aa^\sharp-bb^\sharp=AA^\sharp-BB^\sharp=1.
$$
Put $d=aA^\sharp-bB^\sharp$. Select bounded analytic columns and divide by their determinants to define
$$
M=\begin{cases}([\mu_2]_1/a,[\mu_3]_2),&k\in D_1,\\([\mu_1]_1/d,[\mu_3]_2),&k\in D_2,\\([\mu_3]_1,[\mu_1]_2/d^\sharp),&k\in D_3,\\([\mu_3]_1,[\mu_2]_2/a^\sharp),&k\in D_4.\end{cases}
$$
For example, $[\mu_3]_2=b e^{-2i\theta}[\mu_2]_1+a[\mu_2]_2$, so the first pair has determinant $a$ before division. The corresponding determinant for the second pair is $d$. The other two are their reflected counterparts. Hence $\det M=1$ and $M\to I$ at spectral infinity.
Here are explicit connection matrices furnishing every jump. Write $E=e^{-2i\theta}$; wherever the boundary values coexist, $M_j=\mu_2C_j$, with
$$
C_1=\begin{pmatrix}1/a&bE\\0&a\end{pmatrix},\quad C_2=\begin{pmatrix}A^\sharp/d&bE\\B^\sharp E^{-1}/d&a\end{pmatrix},\quad C_3=\begin{pmatrix}a^\sharp&BE/d^\sharp\\b^\sharp E^{-1}&A/d^\sharp\end{pmatrix},\quad C_4=\begin{pmatrix}a^\sharp&0\\b^\sharp E^{-1}&1/a^\sharp\end{pmatrix}.
$$
Orient each ray away from zero; the plus side is its left side. Then $M_+=M_-J$, with \b[$J=C_-^{-1}C_+$], entirely determined by the spectral functions. For instance, on the positive real axis,
$$
J=C_4^{-1}C_1=\begin{pmatrix}1/(aa^\sharp)&(b/a^\sharp)e^{-2i\theta}\\-(b^\sharp/a)e^{2i\theta}&1\end{pmatrix}.
$$
The relevant <Riemann-Hilbert problem> contour is therefore
$$
\boxed{\Sigma=\mathbb R\cup i\mathbb R.}
$$
It is a cross because time analyticity involves the sign of $\operatorname{Im}k^2$, not just $\operatorname{Im}k$. Zeros of the denominators require residue conditions or small contour detours; they may not simply be ignored. The <half-line NLS Riemann-Hilbert reconstruction> recovers $q=2i\lim_{k\to\infty}kM_{12}$, by matching the constant term $i[\sigma_3,M_1]=Q$ in the spatial <Lax pair>.
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