This answer addresses Q2, as Q1 was more or less answered in the comments. Edit of Oct 21, 2017, an upper bound and a big correction.
You can get a lower bound on the area by the fact that the diameter (in the sense of metric spaces) of the starting square must be greater than or equal to the diameter of the final embedded torus with its intrinsic metric.
I think the diameter of the standard embedded torus is realized by the distance between two points on the outer equator that are opposite each other.
I was able to extract the following from the note Geodesics on the Torus
and other Surfaces of Revolution
Clarified Using Undergraduate Physics Tricks
with Bonus: Nonrelativistic and Relativistic
Kepler Problems by Robert Jantzen.
For a standard torus with major radius 2 and minor radius 1 (what Jantzen calls the "unit ring torus" and what I'll denote $T_{emb}$), Jantzen computes that the shortest path connecting two antipodal points on the outer equator is of length approximately 15.26/2 = 7.63. Below is his picture (Fig. 11) of the "[1,1;0]" geodesic that connects them:
The path is depicted in blue. Somewhat confusingly, a meridian is also colored in blue in this image; that should be disregarded.
The diameter of an open square of area $a^2$ is $\sqrt{2}a$, realized by a diagonal joining two corners.
The square with diameter 7.63 has side length $a=5.39$ and area 29.1, which gives a lower bound on the area of a wrapping square.
This is about 1.47 times the area of the unit ring torus $T_{emb}$ $Rr\pi^2=2\pi^2\approx19.739$ (formula from MathWorld).
Now, this is a pretty awful bound since it's intuitively obvious that if we try to wrap $T_{emb}$ with a square so that the diagonal of the square lies along the above-depicted diameter, there's a lot of the torus that won't be covered (more than half, it seems). This does beat the trivial "area" bound though.
I now consider a different problem: that of finding a map from the flat square torus with "side length" $a$, $T_{flat}(a)$, to $T_{emb}$ which is a wrapping in the sense of the question.
Note that every wrapping from $T_{flat}(a)$ to $T_{emb}$ induces wrapping maps from a square with side length $a$ to $T_{emb}$, so it is possible to get upper bounds for the area of the smallest wrapping square by constructing a wrapping from a flat torus.
For instance, we can get an upper bound by the following obvious map from $T_{flat}(a)$ to $T_{emb}$: let the square fundamental domain of $T_{flat}(a)$ be parametrized by $(x,y)\in[0,a]^2$. Then we map this to the embedded torus so that constant $x$ circles go to longitudinal circles and constant $y$ circles go to meridians.
For what values of $a$ is this map a wrapping? All meridians have the same length $2\pi$ which means that we must take $a\geq 2\pi$ for a wrapping. The longest longitudinal circle is the outer equator of the torus, which has length $6\pi$, which means that in fact we must take $a\geq 6\pi$. One can check that this is also sufficient for the map to be a wrapping (by computing the Jacobian of the above map and seeing that its operator norm $\leq1$ precisely when $a\geq 6\pi$). Therefore:
The above map is a wrapping from $T_{flat}(6\pi)$ to $T_{emb}$ and so an upper bound for the area you're looking for is $36\pi^2\approx 355.3$, which is 18 times the area of $T_{emb}$.
Unfortunately, results relating to this problem don't obviously give lower bounds to Q2. This is because wrappings from a square to $T_{emb}$ don't have to "glue up" to maps from $T_{flat}(a)$ to $T_{emb}$. [In earlier versions of this answer, I did not realize this!]
Nonetheless, I think this is an independently interesting problem (maybe a worthy "Q3"?) and perhaps the results below inspire some other ideas. In particular, I find it unlikely that the best wrapping surjection from a square has a smaller area than the lower bound given for $T_{flat}(a)$ below, though I don't know how to prove it. Indeed, I would be surprised to see any wrapping that does better than the one given as an upper bound above!
In what follows I'll discuss a lower bound for wrappings from $T_{flat}(a)$ to $T_{emb}$ using the idea described in the first part of this answer. The diameter $T_{flat}(a)$ is $a/\sqrt{2}$, realized by a point in the "center" of the square fundamental domain and a vertex.
The flat square torus with the same diameter as $T_{emb}$ has $a=10.79$ and area 116.4, which gives a lower bound on the area of a wrapping flat torus.
This is about 5.90 times the area of $T_{emb}$, $Rr\pi^2=2\pi^2\approx19.739$.
This bound is very unlikely to be sharp for this problem. Note that the diameter of $T_{flat}(a)$ is realized by 4 paths (the segments of the "crossed diagonals"). See the green paths in this figure, which is an edit of this file from wikipedia.
The diameter of $T_{emb}$ is also realized by 4 paths (the "top" and "bottom" of the [1,1;0] geodesic depicted above, as well as the reflections of these through the $xy$ plane). A crude cartoon is depicted below, where the green paths represent the geodesics (but were not actually computed as such).
However, the graphs formed by the union of these 4 paths in the two spaces have topologically inequivalent embeddings in their respective tori! One way to see this is to consider the (topological) circles formed by the edges in the graphs. For the graph in $T_{flat}(a)$, such a circle never bounds a disk, but it can (in multiple ways) for the graph in $T_{emb}$.
In the following figure, the left panel shows the previous sketch of the diameters on $T_{emb}$ with a red longitudinal circle and blue meridional circle highlighted. If we imagine cutting $T_{emb}$ along those circles and unwrapping it, the graph of diameters will be isotopic to the one in the figure in the right panel.
I don't know how to show that this precludes the existence of a wrapping of this size but it is certainly boggling my mind.
