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.
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 $$ \|u - v\|_{H^1_0(\Omega)} \leq C \| \frac{\nabla u}{|\nabla u|} - \frac{\nabla v}{|\nabla v|} \|_{L^2(\Omega)}$$
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.