= Solution
Use the negative-exponential <Half-range Fourier transform>
$$
\widehat q(k,t)=\int_0^\infty e^{-ikx}q(x,t)\,dx,\qquad F(k)=\widehat q(k,0),\qquad \operatorname{Im}k\leq0.
$$
For the boundary traces $g_j(t)=\partial_x^jq(0,t)$, define <finite-time spectral boundary transforms> $G_j(k,t)=\int_0^t e^{w(k)s}g_j(s)\,ds$. Integrating the conservation law in $x$ and then time gives the <global relation>
$$
e^{w(k)t}\widehat q(k,t)=F(k)+G_2(k,t)+ikG_1(k,t)-(k^2+1)G_0(k,t).
$$
The plus sign of the boundary contribution follows because the integrated flux is $-E(0,t)X(0,t,k)$.
<Fourier inversion> consequently gives a real-line initial integral plus a boundary integral. To put the latter in the complex plane, define
$$
D_+=\{k=u+iv:v>0,\ \operatorname{Re}w(k)<0\}=\{u+iv:v>\sqrt{3u^2+1}\}.
$$
Orient $\partial D_+$ from the left upper infinity, through $i$, to the right upper infinity, with $D_+$ on the left. Each $G_j$ is entire, and in the region between the real axis and this contour, $\operatorname{Re}w\geq0$. The causal factors $e^{-w(t-s)}$ are bounded there and $e^{ikx}$ decays for $x>0$. The <Cauchy integral theorem>, with the usual large-arc limiting argument, deforms the boundary integral to give
$$
\boxed{q(x,t)=\frac1{2\pi}\int_{\mathbb R}e^{ikx-wt}F(k)\,dk+\frac1{2\pi}\int_{\partial D_+}e^{ikx-wt}\bigl[G_2+ikG_1-(k^2+1)G_0\bigr]dk.}
$$
Initial smoothness, spatial decay and corner compatibility justify inversion; if necessary the integrals are first regularized before contour deformation. The unknown Neumann-type traces $g_1,g_2$ will be removed by the <global relation>.
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