The answer to the first is NO.
As mentioned by another answer you can use Do Carmo to prove a local result of the type: at points where the normal curvature is non-zero the 0-directions defines a foliation of straight lines.
With this one can prove that embedding $\mathbb{R}²$ into $\mathbb{R}^3$ preserves the diameter of circles, but one can construct a counter example in the case of a piece of paper:
Take a equilateral triangle on your table with points barely out side of the piece of paper. Then since these sides do not meet inside the piece you may fold (using very sharp foldings) the piece of paper up at the sides of the triangle such that out side of a nighborhood of the triangle the paper lies in planes which are perpendicular to the table. Note that this can not be extended to all of $\mathbb{R}^2$ because the folding lines interset! Then shrinking the circumscribed sphere slightly of the triangle to lie in the paper - this is folded up so that the diameter becomes strictly less.
For the second question I really dont see $\pi_1$ any where. What I see is the following: for me it is natural to defined the distance on a Riemmanian manifold $X$ as:
$d(x,y) = \inf_{\gamma} \textrm{len}(\gamma)$
where the infimum is over curves $\gamma$ from $x$ to $y$, and len($\gamma$) is the length of the curve. This is intrinsic and gives the same diameter diameters for the piece of paper, but when you removed remove an open set it becomes different, because the curves has to avoid this set.
Since isometric embeddings preserve lengths of curves you get for my with this definition
$d(Fx,Fy) \leq d(x,y)$
