The reasoning would be exactly the same as in the case of the electric field. The formula you wrote is valid only within the material.
In the electric case, when a material is polarized with dipole moment per unit volume $\mathbf{P}$ then two bound charge densities appear:
$$\sigma_b = \mathbf{P} \bullet \mathbf{\hat{n}}$$
$$\rho_b = -\nabla \bullet \mathbf{P} $$
But when using Gauss's law within the material, we write
$$\epsilon_0\nabla\bullet \mathbf{E} = \rho_b + \rho_f$$
Perhaps this is easier if we look at this equation in its integral form:
$$\iint \limits_{S}\mathbf{E}\bullet \mathrm{d}\mathbf{S}=\frac{Q_{total}}{\epsilon_0}$$
The surface bound charge density could be represented as a volume density if we write it like
$$\rho_{b_{surface}}=\delta(\mathbf{r}-\mathbf{r}_0)\sigma_b$$
where $\mathbf{r}_0$ is the location of the surface in whatever coordinates you want. So now it's easy to see that when $S$ (our surface of integration) does not include the surface of the dielectric, the delta function doesn't add anything to our field. However, if you want to find the field outside the material, then you have to integrate that delta (and you will be left with an added $\sigma_b$, just as expected).
I wrote the answer based on the electric case because IMHO it is easier to understand it that way. In the magnetic case the reasoning is exactly the same. The surface current would occupy the role of the surface charge, and we don't need that to find the field inside our material.
