= Solution
At $p$, the surface $\widetilde S$ cannot cross the boundary $S=\partial\mathcal R$, because it lies inside $\mathcal R$. Consequently the two surfaces are tangent at $p$. Apply a <proper Euclidean motion of Euclidean three-space> so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward <unit normal> is $N=e_3$. Locally write
$$
S:\ z=f(x,y),
\qquad
\widetilde S:\ z=g(x,y),
$$
with the region $\mathcal R$ on the side $z\geq f(x,y)$. Then
$$
g\geq f,
\qquad
g(0)=f(0),
\qquad
\nabla g(0)=\nabla f(0)=0.
$$
Thus $g-f$ has a <local minimum> at zero and its <Hessian matrix> is <positive semidefinite matrix>[positive semidefinite]. The <fundamental forms of a graph surface> at a horizontal tangent plane give
$$
H(0)=\frac12\operatorname{tr}\operatorname{Hess}f(0),
\qquad
\widetilde H(0)=\frac12\operatorname{tr}\operatorname{Hess}g(0).
$$
Taking the trace of the positive-semidefinite difference proves the <mean-curvature comparison at tangential contact>:
$$
\boxed{\widetilde H(p)\geq H(p).}
$$
There is no analogous conclusion for $K$. In the same local coordinates take
$$
f(x,y)=-x^2-y^2,
\qquad
g(x,y)=-\frac12(x^2+y^2).
$$
Then $g\geq f$, so the second graph lies on the inward side of the first, while the <Gaussian curvature of a graph surface> gives
$$
H_f(0)=-2,
\qquad H_g(0)=-1,
\qquad
K_f(0)=4,
\qquad K_g(0)=1.
$$
Hence the mean-curvature inequality holds but
$$
\boxed{\widetilde K(p)<K(p).}
$$
Using smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the <Gaussian curvature has no tangential-contact comparison principle>[counterexample] under the global hypotheses.
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