I want to know if the angular momentum equation $\frac{dL}{dt}=\vec\tau$ holds for the spherical momentum. Suppose that the angle $\theta$ is constant and the mass goes on uniform circular motion around the rotation axis. Then $\vec L =\vec L_z$ is conserved. However $\vec \tau=\vec r \times m\vec g \neq 0$. Why is it so? 
More generally, I would like to know when the angular momentum equation applies, and why it does or doesn't apply in this specific case.
You're conflating the vector $\vec{L}$ with one of its components $L_z$.
The torque on the pendulum is in fact non-zero, which means that the vector $\vec{L}$ is not constant, according to $$ \vec{\tau} = \frac{d\vec{L}}{dt}. $$ However, since $\vec{g}$ is always in the $z$-direction, the component of $\vec{\tau}$ along the $z$-axis must be zero. (This is because $\vec{\tau} = m \vec{r} \times \vec{g}$, and so $\vec{\tau}$ will be perpendicular to both $\vec{r}$ and $\vec{g}$.) Taking the $z$-component of the above equation, this means that $$ \tau_z = \frac{dL_z}{dt} = 0 $$ and so $L_z$ is in fact a constant.
The other two components of the angular momentum ($L_x$ and $L_y$) will vary with time, since neither $\tau_x$ nor $\tau_y$ will be zero in general. This means that the whole vector $\vec{L}$ isn't a constant, since two of its components will be changing. It's only $L_z$ that will stay constant.
The definition of torque is $\vec \tau \equiv \dfrac{d\vec L}{dt}$. That equation will always hold (in fixed inertial frame/COM frame), as it is what torque is defined to be.
I'd suggest you visit When does torque equal to moment of inertia times the angular acceleration?. It may clear up your confusion.
As for this particular case, getting a non zero value for torque is correct -- look at the angular momentum of the pendulum bob at different point on the trajectory -- notice how the direction of the angular momentum vector will not stay the same (recall that angular momentum vector is perpendicular to your linear momentum and $\vec l$). The torque by gravity will change the angular momentum.
And no, there being torque does not automatically imply angular acceleration. Euler's second law states that,
$$\vec \tau = \vec r_\mathrm{CM} \times \vec a_\mathrm{CM}m+I\vec \alpha$$ Therefore, $\alpha$ can be zero, and you can still have a net torque.
EDIT: the reason in lots of intro physics courses the $\tau=I\alpha$ equation is present, is because in those classes, $a_\mathrm{cm}=0$, so the first term of Euler's second law is zero.
lets obtain the torque about the z- axis.
$$\tau=m\,v\,r+ \underbrace{\left(\vec r\times m\,\vec g\right)\cdot \vec e_z}_{=0}\tag 1$$
with:
$$r=l\sin(\theta)~,v=l\,\dot\varphi\,\sin(\theta)$$ $$\vec r=\begin{bmatrix} r\cos(\varphi) \\ r\sin(\varphi) \\ 0 \\ \end{bmatrix}~,\vec g=\begin{bmatrix} 0 \\ 0 \\ -g \\ \end{bmatrix}$$
$\Rightarrow$
$$\tau=m\,l^2\,\sin^2(\theta)\dot{\varphi}\tag 2$$
from the equation of motion with $~T=\frac 12 m\,v^2~,U=\text{const.}~$ you obtain that
$$m\,l^2\,\sin^2(\theta)\dot{\varphi}=\text{const}=L_\varphi$$
thus the torque $~\tau=\frac{dL}{dt}=L_\varphi~,\Rightarrow L~=L_\varphi\,t~$ is not conserved
If you consider torque and angular momentum relative to the center of the circle, the vertical component of the tension equals the weight and the horizontal component is radial. There is no net torque, and the angular momentum vector along the z axis is constant. If you take your reference point at, O, then then angular momentum vector tilts away from the z axis, the torque from the horizontal force causes its tip to swing around in a horizontal circle. (Apply the right hand rules.)
