Control of the $C^1$ norm of a diffeomorphism

Let $\Omega$ be a smooth open set of $\mathbb{R}^3$ diffeomorphic to the unit ball $B$. Let assumme that the boundary $\partial \Omega=\Sigma$ is also smooth and satisfies:

$$\int_\Sigma H^2 d\sigma \leq C,$$ where $H$ is the mean curvature of $\Sigma$. Does there exists a diffeomorphism for $\Omega$ to $B$ whose $C^1$ is controlled by a cosntant which depends only on $C$.

This question have been tackle in here in the case of a rotationnaly invariant domain, using the fact that in dimension $2$ we control the the diffeomorphism between a ball and a simply conected domain assuming that the curve which bound the domain satisfy a chord-arc condition.

Detail: Let $\Omega$ as above and one diffeomorphism $\psi$ from $\Omega$ to $B$, do we have $$\inf_{\phi\in Diffeo(B)} \vert \nabla (\phi\circ \psi)\vert \; \leq K,$$ where $K$ depends only on $C$.

In fact, i am also searching a reference for the problem in dimension $2$ (perhaps here we can replace diffeomorphism by conformal diffeomorphism) replacing $\int_\Sigma H^2 d\sigma \leq C$ by $\int_\Gamma \kappa^2 d\sigma \leq C$ where $\Gamma=\partial \Omega$.

I am ok if "global geometric" hypothesis are needed but only "global geometric"( i.e. about volume, area, length, total curvature).

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Below you say something about searching for the "best" diffeomorphism but in your question you say nothing about this. Also, no definition of "best" is provided. –  Deane Yang Feb 7 '13 at 13:06

Basically, you are asking if $L^\infty$ bounds for first derivatives can be controlled by $L^2$ bounds for second derivatives. This works in one dimension, but not two (Sobolev imbedding theorem).
In general yes, But here we search the best diffeomorphism from $\Omega$ to $B$. Unless you have a counterexample? –  Paul Feb 5 '13 at 16:28