After further examining the original question, and the source for the question, which was in the book "Electromagnetic Theory" by Ferraro on p.543, I was able to understand the conserved quantity $\vec{J}$ as thus.
Considering that with $\vec{B}(\vec{r}) = \frac{g\vec{r}}{r^3}$ the Lorentz force yields:
$\begin{align}
m\ddot{\vec{r}} &= q\dot{\vec{r}} \times \vec{B}\\
\vec{r}\times m\ddot{\vec{r}} &= q[\vec{r}\times(\dot{\vec{r}} \times \vec{B})]\\
\vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\vec{r}\cdot\vec{B}) - \vec{B}(\vec{r}\cdot \dot{\vec{r}})]\\
\vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\frac{g}{r}) - \vec{B}(r\dot{r})]\\
\vec{r}\times m\ddot{\vec{r}} &= q[\dot{\vec{r}}(\frac{g}{r}) - \vec{r}(\frac{\dot{r}g}{r^2})]\\
\vec{r}\times m\ddot{\vec{r}} &= qg\Big[\frac{\dot{\vec{r}}}{r} - \vec{r}\frac{\dot{r}}{r^2}\Big]\\
\frac{d}{dt}\Big(\vec{r}\times m\dot{\vec{r}}\Big) &= qg\frac{d}{dt}\Big(\frac{\vec{r}}{r}\Big)\\\\
\therefore \vec{r} \times \vec{p} = qg\frac{\vec{r}}{r} + \vec{J}\\
\end{align}$
Where $\vec{J}$ is an arbitrary constant vector. Thus $\vec{J}$ is conserved in both magnitude and direction, so $J$ is constant.
$\begin{align}
\therefore \vec{J} =\vec{r} \times \vec{p} - eg\frac{\vec{r}}{r} \\
\end{align}$
Thus it follows that
$\begin{align}
\vec{J} \cdot \vec{r} =(\vec{r} \times \vec{p} - eg\frac{\vec{r}}{r} ) \cdot \vec{r} &=
(0) - egr\\
\therefore \vec{J} \cdot \vec{r} &= Jr\cos\theta = -egr\\
\therefore Jr\cos\theta &= -egr\\
\therefore \cos\theta &= -\frac{eg}{J}\\
\end{align}
$
Thus $\theta$ is constant, and $\dot{\theta}$ = 0
