When considering the mechanics of a point particle, the quantity $\nabla \cdot (\mathbf r \times \mathbf p)$ is not well-defined.
The familiar gradient, curl, and divergence operators are objects which act on fields (vector fields in the case of curl and divergence, and scalar fields in the case of the gradient). For example, in Cartesian coordinates,
$$div(A) = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z} $$
$$=\lim_{\epsilon\rightarrow 0} \left[\frac{A_x(x+\epsilon,y,z)-A_x(x,y,z))}{\epsilon}\right] + \ldots $$
Fundamentally, such an operation only makes sense if you can evaluate $A_x(x+\epsilon,y,z)$ and subtract $A_x(x,y,z)$ - in other words, you require $A_x$ to be a quantity which takes some value at every position.
But now consider $\mathbf r \times \mathbf p$, where $\mathbf r = \mathbf r(t)$ and $\mathbf p=\mathbf p(t)$ are the position and momentum of a point-like object at time $t$. How could we apply the divergence operator to this quantity? How does one "evaluate" this vector quantity at neighboring positions and then subtract?
The answer is that you cannot. There is no meaningful sense in which you can calculate a spatial derivative of a vector quantity which is not a field, and so even though $\nabla \cdot (\mathbf r \times \mathbf p)$ may look like a reasonble thing to talk about at first glance, it ultimately is not.
As a side note, if you are talking about the flow of some kind of fluid which has mass density $\rho(\mathbf r)$ and flow velocity $\mathbf u(\mathbf r)$, then the quantity $\boldsymbol \ell = \mathbf r \times (\rho \mathbf u)$ is the angular momentum density of the fluid (calculated with respect to the coordinate origin), and this is a vector field which can be differentiated using the divergence and curl operators. It is rarely spoken of - it is usually much less interesting than the vorticity $\boldsymbol \omega = \nabla \times \mathbf u$ - but it is at least a well-defined quantity.