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The answer is "no" for $n=2$.

Choose $\Gamma=f(\partial M)$ to be a slight variation of flat nonconvex quadrangle. You want to present $\Gamma$ as an intersection of two ruled surfaces which have no parallel tangent planes.

In other words, the set of normal unit vectors formed by two curves (say $\alpha$ and $\beta$) in $S^2$ do not intersect. One of the curves ($\alpha$) is nearly constant, it corresponds to the half of $\partial[\mathop{Conv}\Gamma]$. For the other, choose an involution $\phi$ on $\Gamma$ and connect by line $x$ to $\phi(x)$. The choice can be made so that the obtained ruled surface is smooth. Still we have a lot of freedom to ensure that $\beta$ avoids to go near $\alpha$.

The answer is no for $n=2$. It is sufficient to construct 3 surfaces with common boundary (say $\Sigma_i$, $i\in\{1,2,3\}$) such that there is no choice of points $p_i\in\Sigma_i$ with pairwise parallel tangent planes.

Let us take a smooth function $f:S^1\to \mathbb R$, $f(t)\approx\sin(2\cdot t)$ with one little bump near zero so it has 3 local minima and maxima. We want to construct three functions $h_1,h_2,h_3$ from unit disc $D$ to $\mathbb R$ such that each has $f$ as boundary values and

1. if $\nabla h_1(x)=\nabla h_2(y)$ then $\nabla h_1(x)=0$

2. $\nabla h_3\not=0$ anywhere in the disc.

Then graphs of functions give the needed surfaces. The graphs of $h_1$ and $h_2$ are parts of boundary of convex hull of graph of $f:\partial D\to\mathbb R$; it is easy to check (1).

The graph of $h_3$ is a ruled surface which formed by lines passing through points $(u,f(u))$, $(\phi(u),f(\phi(u))\in\mathbb R^3$, $u\in S^1$ for some involution diffeomorphism $\phi: S^1\to S^1$. To have the property one has to choose   $\phi$ with two fixed points (say at global minima of $f$) so that if $f(\phi(x))=f(x)$ for some $x$ then $f'(\phi(x)\cdot f'(x)<0$. The later is easy to arrange, that is the place we need the bump of $f$.

P.S. Hopefully it is correct now :)

Ups, I just realized that it was wrong, and can no delete accepted answer...

The answer is "no" for $n=2$.

Choose $\Gamma=f(\partial M)$ to be a slight variation of flat nonconvex quadrangle. You want to present $\Gamma$ as an intersection of two ruled surfaces which have no parallel tangent planes.

In other words, the set of normal unit vectors formed by two curves (say $\alpha$ and $\beta$) in $S^2$ do not intersect. One of the curves ($\alpha$) is nearly constant, it corresponds to the half of $\partial[\mathop{Conv}\Gamma]$. For the other, choose an involution $\phi$ on $\Gamma$ and connect by line $x$ to $\phi(x)$. The choice can be made so that the obtained ruled surface is smooth. Still we have a lot of freedom to ensure that $\beta$ avoids to go near $\alpha$.

The answer is "no" for $n=2$. Choose $f(\partial M)$ to be a slight variation of flat nonconvex quadrangle. You may think that $f(\partial M)$ is an intersection of two ruled surfaces which have no parallel tangent planes. (The set of normal unit vectors formed by two curves in $S^2$ and they do not intersect.)

Ups, I just realized that it was wrong, and can no delete accepted answer...