Suppose $\Omega\subset\subset \mathbb{R}^n$, be a bounded, strongly convex domain (with $0\in\Omega$). Is the following inequality true?
$$ C_2\left|x-y\right|^2\leq\left|\frac{x}{|x|}-\frac{y}{|y|}\right|^2\leq C_1\left|x-y\right|^2, \forall x,y\in b\Omega. $$ $C_1,C_2$ are positive constants not depending on $x,y$.
This seems to be true. I don't think it is necessary to assume that $\Omega$ is strongly convex (but convex, open containing the origin, and bounded are all important). Most of the following is just repeating standard arguments about the Minkowski functional.
Let $\mu_{\Omega}$ be the Minkowski functional of $\Omega$, $$ \mu_{\Omega}(x) = \inf \lbrace t > 0 : x \in t \, \Omega\rbrace. $$ The important properties of $\mu_{\Omega}$ we'll need are that it's sublinear, i.e., $\mu_{\Omega}(x + y) \leq \mu_{\Omega}(x) + \mu_{\Omega}(y)$, which follows from the convexity of $\Omega$, and positively homogeneous, i.e., $\mu_{\Omega}(\alpha x) = \alpha \mu_{\Omega}(x)$ when $\alpha \geq 0$. (These properties characterize a sublinear function.)
I'm going to take $x, y \in S_1$, so that the corresponding points in the $\partial \Omega$ are $x/\mu_{\Omega}(x)$ and $y/\mu_{\Omega}(y)$. (Thus my notation does not agree with yours; I found it simpler to work this way.)
First we establish the equivalence of $\mu_{\Omega}$ and $|\cdot|$.
Write $B_1$ to denote the ball of radius $1$ about the origin (in the norm $|\cdot|$). If we let $$ m = \sup \lbrace s: s \, \Omega \subseteq B_1 \rbrace \qquad \text{ and } \qquad M = \inf \lbrace s: B_1 \subseteq s \, \Omega \rbrace, $$ then $m > 0$ since $\Omega$ is bounded, and $M < \infty$ since $\Omega$ is an open neighborhood of the origin. Moreover, for $x \in S^1$ we then have $$ m \leq \mu_{\Omega}(x) \leq M. $$
Since $\mu_{\Omega}$ is positively homogenous, it follows that for all $x \in \mathbb{R}^n$, $$ m |x| \leq \mu_{\Omega}(x) \leq M |x|. $$
We will do the two inequalities separately.
Right inequality.
Lemma. If $\alpha, \beta \geq r \geq 0$ and $x, y \in S_1$ then $|\alpha x - \beta y| \geq r |x - y|$.
Proof. This just says that the distance between two points on the rays through $x$ and $y$ is minimized when both points are as close to the origin as possible. $\square$
By the lemma, for all $x, y \in S_1$ we have $$ \frac{1}{M}|x - y| \leq \left| \frac{x}{\mu_{\Omega}(x)} - \frac{y}{\mu_{\Omega}(y)} \right|, $$ which takes care of the right side of your desired inequality.
Left inequality.
To do the left side, note that, because $\mu_{\Omega}$ is sublinear, $$ \mu_{\Omega}(x) = \mu_{\Omega}(x - y + y) \leq \mu_{\Omega}(x - y) + \mu_{\Omega}(y) $$ and hence $$ \mu_{\Omega}(x) - \mu_{\Omega}(y) \leq \mu_{\Omega}(x - y). $$ Since the same holds with $x$ and $y$ switched, it follows that $$ |\mu_{\Omega}(x) - \mu_{\Omega}(y)| \leq \max \lbrace \mu_{\Omega}(x - y), \mu_{\Omega}(y - x) \rbrace \leq M |x-y|. $$ We estimate \begin{align*} \left| \frac{x}{\mu_{\Omega}(x)} - \frac{y}{\mu_{\Omega}(y)} \right| &\leq \left| \frac{x}{\mu_{\Omega}(x)} - \frac{y}{\mu_{\Omega}(x)} \right| + \left| \frac{y}{\mu_{\Omega}(x)} - \frac{y}{\mu_{\Omega}(y)} \right| \\ &= \frac{1}{\mu_{\Omega}(x)} |x - y| + |y| \left| \frac{\mu_{\Omega}(y) - \mu_{\Omega}(x)}{\mu_{\Omega}(x) \mu_{\Omega}(y)} \right| \\ &\leq \frac{1}{m}|x - y| + \frac{M}{m^2}|x - y| \\ &= \frac{m + M}{m^2} |x - y|, \end{align*} which is the left side of your desired inequality.