Using Noether's Theorem: The transformation $q_i\rightarrow q_i+\epsilon$ is a symmetry iff $\delta L = \dfrac{dF}{dt}$. In that case, the conserved charge is given by: $$\frac{d}{dt}(\frac{\partial L}{\partial \dot q}\delta q-F)=0$$
Here when we do the transformation $\theta\rightarrow\theta+\epsilon$, $F$ comes out to be zero. (You can calculate it using Taylor Expansion) and hence the conserved charge would be: $$\frac{d}{dt}(\frac{\partial L}{\partial \dot\theta}\delta \theta)=0$$ which in this case comes out to be $\dfrac{d}{dt}(mr^2\dot{\theta})$ which implies angular momentum conservation.
