I think you're inadvertently opening a big can of worms. The question can be answered by a combination of two facts: the absence of branch points in (almost-)minimising hypersurfaces and Allard's regularity theorem.

Specifically, the tangent cones to $H$ at $h$ must be multiples of an $n$-dimensional hyperplane $P$ say, with some multiplicity $Q \in \mathbf{Z}_{>0}$. The tangent cones cannot be more complicated minimal cones, as for example a Frankel-type argument demonstrates. Let $$ \mathbf{C} \in \mathrm{VarTan}(H,h)$$ be a tangent cone to $H$ at $h$: this is a (singular) minimal surface. Knowing that $\mathbf{C}$ is a stationary varifold is enough for now. By construction the cone is supported in a closed half-space, for example $$\mathrm{spt} \, \mathbf{C} \subset \{ X \in \mathbf{R}^{n+1} \mid X^{n+1} \geq 0 \}.$$
The intersection of $\mathbf{C}$ with the unit sphere $\partial B$ defines a stationary varifold contained in a hemisphere. Now on the one hand, as $\partial B$ has positive Ricci curvature, $\mathrm{spt} \, \mathbf{C}$ and $\partial B \cap \{ X^{n+1} = 0 \}$ must intersect: this is Frankel's theorem. On the other hand, this intersection must be tangential, and the maximum principle forces them to coincide: $$\mathrm{spt} \, \mathbf{C} = \{ X^{n+1} = 0 \}.$$


Therefore letting $P = \{ X^{n+1} = 0 \}$ one has $$\mathbf{C} = Q \lvert P \rvert.$$


When $H$ is minimising (or almost-minimising), then it cannot have branch point singularities, and necessarily $Q = 1$.

Therefore the tangent cones are multiplicity one tangent planes, and by Allard regularity $H$ must be smooth in a neighbourhood of the point $h$.