Question 1. Yes sure and the same proof should work.
Question 2. The answer is "yes" in $W^{1,p}$ for any $p<\infty$ and "no" in $W^{1,\infty}$.
$W^{1,1}$:
Note that
$$d_{\mathrm{H}}(\Sigma_t,\Sigma_s)<\varepsilon
\ \ \iff\ \
\sup_x\{\,|d_{\Sigma_t}(x)-d_{\Sigma_s}(x)|\,\}<\varepsilon.$$
So it is sufficient to show the following:
If (1) $\sup_x|f_t(x)-f_0(x)|\to 0$ as $t\to 0$,
(2) $|\nabla_xf_t|\le 1$ for any $t,x$ and
(3) $|\nabla_xf_0|=1$ for almost all $x$ then
$$\int\limits_M|\nabla_x f_t-\nabla_x f_0|\cdot dx\to 0.$$
Assume contrary.
Then for some fixed $\varepsilon>0$ and any $t>0$
there is a set $C_t$ of measure $>\varepsilon$ such that $|\nabla_x f_t-\nabla_x f_0|>\varepsilon$ for any $x\in C_t$.
Move (almost) along $\nabla f_0$ and integraite both $df_0$ and $df_1$.
You get a positive lower bound on $\sup |f_t(z)-f_0(z)|$.
The later contradicts that $\sup_x|f_t(x)-f_0(x)|\to 0$.
$W^{1,p}$: All this proves that $f_t$ is continuous in $W^{1,1}$.
Since $\nabla f_t$ is bounded, you get continuity in $W^{1,p}$ for any $p<\infty$.
$W^{1,\infty}$: There is no continuity in $W^{1,\infty}$; this can be seen for $M=\mathbb R$ and $f_t(x)=|x-t|$. (The same works for compact $M$)
It is certainly not true that every complete nonflat open manifold of nonnegative curvature has Euclidean volume growth. Counterexamples are trivial to construct. Say, a capped cylinder. More generally any nonnegatively curved manifold $M^n$ with nontrivial soul has slower than Euclidean volume growth. Because its asymptotic cone at infinity has dimension strictly smaller than $n$ if the soul is not a point while manifolds with Euclidean volume growth have asymptotic cones of dimension $n$.
This also implies that any nonnegatively curved manifold with Euclidean volume growth is diffeomorphic to $\mathbb R^n$.
Best Answer
Consider the complex curve $w = z^k$ in $\mathbb{C}^2\simeq\mathbb{R}^4$, which is calibrated and therefore area-minimizing. The area of the part of this curve that lies inside the polydisk $\max\{|z|,|w|\}\le 1$ is $(k{+}1)\pi$. (The reason is that, because the standard Kähler form $\Omega = \tfrac{i}{2}(\mathrm{d}z\wedge\mathrm{d}\overline{z} + \mathrm{d}w\wedge\mathrm{d}\overline{w})$ calibrates this curve, the total area of this part of the curve is the sum of the areas of the projections, counted with multiplicities, onto the $z$- and $w$-axes. The projection onto the $z$-axis is $1$-to-$1$ onto the unit disk and the projection onto the $w$-axis is $k$-to-$1$ onto the unit disk (except for $w=0$).
Thus, there is no upper bound on the area of an area-minimizing surface inside a fixed ball in $\mathbb{R}^4$.