= Solution
Use a time <Laplace transform> and a spatial <Green function>. The first-order transport term must be retained: the <Airy resolvent kernel> for $\partial_x^3+p$ alone would solve a different equation. The governing <linear partial differential equation> is the <linear dispersive Stokes equation>, distinct from the viscous-fluid <Stokes equation>.
For the time interval of interest, extend the given boundary signal by zero after $T$ and put
$$
G(p)=\int_0^T e^{-ps}g_1(s)\,ds,\qquad \operatorname{Re}p>0.
$$
The possible discontinuity at $T$ does not affect a solution at $t<T$. Equivalently, use any sufficiently regular continuation beyond $T$: the resulting <causal Green function> produces the same earlier solution. We seek the usual solution with spatial decay and a time transform, rather than an unrestricted spatially growing solution on the half-line.
Writing $Q(x,p)$ for the time <Laplace transform>, the <Laplace transform of a derivative> gives
$$
Q_{xxx}+Q_x+pQ=q_0(x),\qquad Q_x(0,p)=G(p).
$$
The corresponding <characteristic roots> are the three <roots of a polynomial>
$$
P_p(r)=r^3+r+p.
$$
For $\operatorname{Re}p>0$, label the unique root of negative <real part> by $r_-(p)$, and the two roots of positive <real part> by $r_1(p),r_2(p)$. This <spatial root splitting for the dispersive Stokes resolvent> is easily checked: if $r=ik$, then $p=-i(k-k^3)$ is imaginary, so no root crosses the imaginary axis as $p$ varies in the right half-plane. For real $p>0$ the cubic is strictly increasing on the real line and has one negative real root, while its conjugate pair has positive <real parts>. The root count remains the same throughout that connected half-plane. A repeated root would satisfy $3r^2+1=0$ and again put $p$ on the imaginary axis. All roots are therefore simple there, and $r_-(p)$ is a nonzero <holomorphic function>.
Set $d_j=3r_j^2+1$, including $j=-$. Define the <Stokes resolvent kernel with advection>
$$
R_p(s)=
\begin{cases}
e^{r_-s}/d_-,&s\ge0,\\
-\displaystyle\sum_{j=1}^2 e^{r_js}/d_j,&s<0.
\end{cases}
$$
The sign on the negative half-line is essential. To verify the <Green function>, use the <partial fraction decomposition>
$$
\frac1{P_p(r)}=\sum_{j\in\{-,1,2\}}\frac1{d_j(r-r_j)}.
$$
Its expansion at infinity yields the <reciprocal-polynomial root identities>
$$
\sum_j\frac1{d_j}=0,\qquad
\sum_j\frac{r_j}{d_j}=0,\qquad
\sum_j\frac{r_j^2}{d_j}=1.
$$
Consequently $R_p$ and $R_p'$ are continuous at zero, and $R_p''$ has jump one. Each piece solves the homogeneous <ordinary differential equation>, so the <jump conditions for a third-order Green function> imply, in the sense of <distributional derivatives>,
$$
(p+\partial_s+\partial_s^3)R_p=\delta_0.
$$
Both pieces decay in their respective spatial directions.
Now define
$$
V(x,p)=\int_0^\infty R_p(x-y)q_0(y)\,dy.
$$
It satisfies the transformed inhomogeneous equation. The difference $Q-V$ solves the homogeneous <linear ordinary differential equation>; spatial decay leaves only $e^{r_-x}$. Imposing the <Neumann boundary condition> fixes its coefficient, giving the <half-line Stokes solution with a boundary derivative>
$$
Q(x,p)=V(x,p)+\frac{e^{r_-x}}{r_-}\bigl[G(p)-V_x(0,p)\bigr].
$$
No division by a vanishing root is involved, since $r_-=0$ would require $p=0$. Moreover the remaining boundary trace is explicitly known from $q_0$:
$$
V_x(0,p)=-\sum_{j=1}^2\frac{r_j}{d_j}
\int_0^\infty e^{-r_jy}q_0(y)\,dy.
$$
Thus introduce the completely explicit initial-data kernel
$$
K_p(x,y)=R_p(x-y)-\frac{e^{r_-x}}{r_-}R_p'(-y)
=R_p(x-y)+\frac{e^{r_-x}}{r_-}
\sum_{j=1}^2\frac{r_j e^{-r_jy}}{d_j}.
$$
For any upward <Bromwich contour> $\operatorname{Re}p=\sigma>0$ to the right of any transform singularities, the requested <integral representation> is
$$
\boxed{q(x,t)=\frac1{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}
e^{pt}\left[\int_0^\infty K_p(x,y)q_0(y)\,dy
+\frac{e^{r_-(p)x}}{r_-(p)}\int_0^T e^{-ps}g_1(s)\,ds\right]dp.}
$$
The contour integral is understood in the standard <Bromwich inversion formula> sense, with truncation limits if it is not absolutely convergent. Every term is expressed in the supplied data and algebraic <characteristic roots>; there are no unknown Dirichlet or second-derivative boundary traces.
The transformed equation verifies the <linear partial differential equation> and the initial term, while taking the spatial <derivative> gives $Q_x(0,p)=G(p)$. <Bromwich inversion formula> therefore supplies the prescribed derivative for $0<t<T$ and the initial value for $x>0$. For zero initial and boundary data, the transformed solution is a decaying homogeneous mode with zero boundary derivative, hence vanishes; this proves uniqueness in the stated transform/decay class. A change of the boundary continuation after $T$ has transform $e^{-pT}$ times another causal transform and contributes only after $T$, by the <Laplace transform time-shift rule>. The printed dot on $q_0$ is interpreted as its derivative in its sole variable $x$, so the stated corner compatibility is $q_0'(0)=g_1(0)$. Additional differentiability at the corner requires the corresponding higher-order compatibility, but is not needed for the representation at positive $x,t$.
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