The sphere is not uniformly close to being isometric to Euclidean space

Here is a simple answer for a simpler question: For every domain $V\subset R^n$, there exist an $\epsilon>0$ such that there is no diffeomorphism $f:U→V$ satisfying $\text{dist}(df,\text{SO})< \epsilon$ everywhere on $U$.

Proof: assume otherwise, then you have a sequence $f_k:U\to V$ such that $\text{dist}(df_k,\text{SO})< 1/k$. This implies that the local Lipschitz constant of $f_k$ is $L(f_k)< 1+ C/k$ (for some $C$ depending only on the dimension). By Arcela-Ascoli you get that $f_k$ converges uniformly to $f:U\to V$ and that $L(f)\le 1$. A simple topological argument gives you that $f$ is surjective. The same arguments work for $f_k^{-1}$, converging to a surjective $g:V\to U$ with $L(g)\le 1$. This implies that $U$ and $V$ are isometric (see Theorem 1.6.15 in Burago-Burago-Ivanov).

Maybe one can improve this proof into what you want...

Note that condition $\text{dist}(df,\text{SO})< \epsilon$ is equivalent to the purely metric conditions (i) $L(f) < 1+\epsilon$, (ii) $f$ is locally invertible and (iii) $L(f^{-1})< 1+ \epsilon$, where $L$ is the local Lipschitz constant. The argument above therefore works for any two non-isometric compact metric spaces $U,V$.

$U$ can be taken to be a metric ball of radius $\delta$ around $s\in U.$ I claim:

1. The image of $f$ includes at least a ball of radius $\delta c_1(\epsilon)$ around $f(s)$ where $c_1(\epsilon)\to 1$ as $\epsilon\to 0.$

   When $\text{dist}(df,\text{SO})$ is small, path lengths are approximately preserved; in particular there is some inequality of the form $|\gamma'|\leq c_1(\epsilon)^{-1}|(f\circ \gamma)'|$ for paths $\gamma:[0,1]\to U,$ with $c_1(\epsilon)\to 1$ as $\epsilon\to 0.$ Given a straight-line path of length at most $\delta c_1(\epsilon)$ starting at $f(s),$ we can lift it to a path of length at most $\delta$ in $U$ starting at $s.$

   To perform this lift we just need the solution of the ODE $df_{\gamma(t)}\circ \gamma'(t)=\text{const}.$ This exists by Picard–Lindelöf, shrinking $\delta$ if necessary to ensure that the second derivative of $f$ is bounded.

2. The volume of the image of $f$ is at most $c_2(\epsilon)\operatorname{vol}(U)$ where $c_2(\epsilon)\to 1$ as $\epsilon\to 0.$

   I mean the usual Riemannian volume, $\sqrt{|g|}dx_1\wedge \dots \wedge dx_n$ in co-ordinates. This definition (or whatever other definition you prefer) shows the volume form is controlled by $g.$

But 1 and 2 are in contradiction for small $\epsilon$ because $\operatorname{vol}(U)$ is smaller than the volume of a Euclidean $\delta$-ball.