Considering you are aware of conservation of total angular momentum in a sphere (if not, I will prove it below), from the lagrangian I think you are using you get:
$$\mathcal{L}=\dfrac{1}{2}R^2\left(\dot\theta^2+\sin^2\theta\,\dot\phi^2\right)$$
$$l_\theta=\dfrac{\partial\mathcal{L}}{\partial\dot\theta}=\dot\theta$$
$$l_\phi=\dfrac{\partial\mathcal{L}}{\partial\dot\phi}=\sin^2\theta\,\dot\phi=const\quad \left(\text{since }\dfrac{\partial\mathcal{L}}{\partial\phi}=0\right)$$
for $r=R\,$ fixed for a sphere of radius $R$. You can see $l_\theta$ and $l_\phi$ are the conjugated momenta associated to $\theta$ and $\phi$, respectively.
The total angular momentum of the system $L$ obeys the following:
$$L^2=mR^2\left(\dot\theta^2+\sin^2\theta\,\dot\phi^2\right)$$
Defining $l^2=\dfrac{L^2}{mR^2}$ and using what we found above: $\,\dot\theta^2=l_\theta^2\quad\text{and}\quad\,\dot\phi^2=\dfrac{l_\phi^2}{\sin^4\theta}$.
Thus:
$l^2=l_\theta^2+\dfrac{l_\phi^2}{\sin^2\theta}$
We would like to show that this total angular momentum is conserved as well. Noting that differentiating respect to a parameter $\lambda$ we get this is conserved for the curve parametrized by $\lambda$:
$$\dfrac{d\,l^2}{d\lambda}=\dfrac{d}{d\lambda}\,\dot\theta^2+\dfrac{d}{d\lambda}\left(\dfrac{l_\phi^2}{\sin^2\theta}\right)=2\left(\ddot\theta-\dot\phi^2\sin\theta\cos\theta\right)\dot\theta=0$$
because the result involves the equation of motion for $\theta$ you already computed, when is equal to zero.
Furthermore,
$$\dot\theta=\sqrt{l^2-\dfrac{l_\phi^2}{\sin^2\theta}}$$
$$\dot\phi=\dfrac{l_\phi}{\sin^2\theta}$$
and from here you can as well try to integrate both equations separately. My recommendation would be trying to find $\phi=\phi(\theta)$, so for instance you can make:
$$\dot\phi=\dfrac{d\phi}{d\theta}\dot\theta$$
What's more:
$$\dfrac{d\phi}{d\theta}\sqrt{l^2-\dfrac{l_\phi^2}{\sin^2\theta}}=\dfrac{l_\phi}{\sin^2\theta}$$
$$\Rightarrow\quad\dfrac{d\phi}{d\theta}=\dfrac{l_\phi}{l}\dfrac{1}{\sin\theta\sqrt{\sin^2\theta-\left(\frac{l_\phi}{l}\right)^2}}$$
Finally, integrating respect to $\theta$ leads to :
$$\phi(\theta)=\phi_0+\arctan\left(\dfrac{\frac{l_\phi}{l}\cos\theta}{\sqrt{\sin^2\theta-\left(\frac{l_\phi}{l}\right)^2}}\right)$$
You can use any plotter you know for seeing how this can give you portions of arc of a sphere (parallels and meridians, e.g.) for a Parametric 3D Plot, setting $(r=R,\theta\in[0,\pi],\phi=\phi(\theta))$. You can get, for example, the equator for $l_\phi=0$.