The total angular momentum of the disk doesn NOT change. There is an outflow of angular momentum due to the shearing of matter rotating at different velocities. Faster inner annuli get a spin down torque from slower outer ones, so they lose angular momentum in favour of the angular of momentum of the outer annuli.
I can express this better and answering the second question with the equation of the angular momentum. Consider an annulus with internal radius $R$ and extension $\Delta R$, in your notation the angular momentum of the annulus is $2\pi R\Delta R \Sigma \, R^2\Omega$. Its variation is given by the advection of mass with different angular momentum in the annulus from both outer and inner annuli plus the viscous torque exerted by the shearing of our annulus with the neighboring inner and the outer annuli. We then get
$$\begin{split}
\frac{ \partial }{ \partial t} \bigl(2 \pi R \, \Delta R \, \Sigma \, R^{2} \Omega \big)\, & = \, u_{R}(R,t) \cdot 2 \pi R \cdot \Sigma (R,t) \cdot R^{2} \Omega \, \\
& - \, u_{R}(R+ \Delta R,t) \cdot \, 2 \pi (R+ \Delta R) \cdot \Sigma (R+ \Delta R,t) \cdot (R+ \Delta R)^{2} \Omega \, \\
&+ \, G(R+ \Delta R,t) \, - G(R,t)
\end{split}
$$
Where $u_{R}$ is the radial velocity and $G(R,t)$ is the torque of an outer annulus acting on a neighboring inner one at radius R.
Taking the limit for $\Delta R \rightarrow 0$ we get
$$
\frac{\partial} {\partial t} \bigl(\Sigma R^{2} \Omega \bigr)\, +\, \frac{1}{R} \frac{\partial} {\partial R} \bigl( R \Sigma u_{R} R^{2} \Omega \bigr)\,=\,
\frac{1}{2\pi R} \frac{\partial G}{\partial R}
$$
So, summarizing:
The first and the second terms on the left hand side are respectively the rate of change of angular momentum per unit area of an annulus at R and the net rate of angular momentum loss from this unit area due to advection of angular momentum with the radial flow. Any imbalance between these two terms implies that the angular momentum content of this region is evolving as result of a net (viscous) torque on the region which is represented by the right hand side.
If you were wondering about the functional form of this torque it is $G(R,t) \, = \, 2 \pi R \, \nu \Sigma R^{2} \frac{d\Omega}{dR}$.
In the case of Keplerian rotation(that is the $\Omega$ you wrote), we can rewrite the angular momentum equation making explicit the torque:
$$
\frac{\partial} {\partial t} \bigl(\Sigma R^{2} \Omega \bigr)\, +\, \frac{1}{R} \frac{\partial} {\partial R} \bigl( R \Sigma u_{R} R^{2} \Omega \bigr)\,=\,
\frac{1}{R} \frac{\partial}{\partial R} \Biggl( \nu \Sigma R^{3} \frac{d \Omega}{dR} \Biggr)
$$
