Let $\Omega$ be the unit ball in $\mathbb{R}^2$. 

Let $u_k(x,y) = \tan^{-1}(k^3 x)$. 

Let $v_k(x,y)$ be a function that agrees with $u_k$ on $\partial\Omega$, and is constant on the level sets of $\{ (x- k)^2 + y^2\}$. 

So $\nabla u_k / |\nabla u_k| = \partial_x$, and 
$ \nabla v_k / |\nabla v_k| = \partial_x + O(1/k) $, so the directions of their gradients differ only by a little. 

Let $B_k$ be the ball of radius $1/k^3$ centered at $(k - \sqrt{k^2 + 1} \approx -1/(2k), 0)$. 

For large $k$, we have that $\nabla u_k \approx 1/k$ on $B_k$. 

On the other hand, $\nabla v_k \approx k^3$ on $B_k$. 

This implies that $\|u_k - v_k\|_{H^1_0(\Omega)} \geq \|\nabla u_k - \nabla v_k\|_{L^2(B_k)} \approx 1$ is bounded uniformly away from zero. 

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This shows that, under the assumptions that

- $\Omega$ is simply connected with connected boundary, and bounded
- $u,v\in C^1(\Omega)$ are both functions bounded by $M$
- $u = v$ on $\partial\Omega$. 
- and that $\nabla u, \nabla v \neq 0$ on $\Omega$

there **does not** exist a constant $C$ such that 
$$ \|u - v\|_{H^1_0(\Omega)} \leq C \| \frac{\nabla u}{|\nabla u|} - \frac{\nabla v}{|\nabla v|} \|_{L^2(\Omega)}$$

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_Edit_: let me give a slightly easier to check counterexample. 

Let $u_k = \tan^{-1}( k^3 (\sqrt{(x-k)^2 + y^2} - k) )$

Let $v_k = \tan^{-1}( k^3 (\sqrt{(x-k)^2 + 1 - x^2}-k))$

Along the set $\{x^2 + y^2 = 1\}$ to two functions obviously agree. 

Now let $B_k$ be the ball of radius $k^{-3}$ centered at the origin. 

The gradients of the functions can be computed entirely explicitly

$$ \nabla u_k = \frac{k^3}{1 + k^6\left(  \sqrt{(x-k)^2 + y^2} - k \right)^2 }
     \frac{1}{\sqrt{(x-k)^2 + y^2}} \cdot (x-k, y) $$

$$ \nabla v_k = \frac{k^3}{1 + k^6 \left( \sqrt{(x-k)^2 + 1 - x^2} - k \right)^2} \frac{1}{\sqrt{(x-k)^2 + 1 - x^2}} (-k,0) $$

Evaluating at the origin one finds

$$ \nabla u_k(0,0) = (-k^3, 0) $$

and

$$ \nabla v_k(0,0) = \frac{k^3}{1 + k^6 (\underbrace{\sqrt{1 + k^2} - k}_{\approx k^{-1}})^2} \frac{1}{\sqrt{1+k^2}} (-k,0) = O(k^{-1})$$ 

and the argument proceeds similarly to above.