One method can be found in Toponogov's book *Differential geometry of curves and surfaces* as suggested by Liviu Nicolaescu in the comments.
Let me adapt the proof given in problem 1.7.10 in the book to the present situation.

Consider the circle $C_\epsilon$ of curvature $\kappa-\epsilon$ instead.
If we can show that $\partial\Omega$ does not meet $C_\epsilon$ for any $\epsilon>0$, we get the desired claim in the limit $\epsilon\to0$.

For simplicity, let us denote the circle by $C$ and assume it has constant curvature $k<\kappa$.
Parametrize $\gamma$ by arc length $s$ starting from $x$, and denote the curvature function by $\kappa(s)$.
Also parametrize $C$ by arc length in the same direction.
Let $\alpha_\gamma(s)$ and $\alpha_C(s)$ be the angles of the tangent compared to those at $x$ (that is, $s=0$).
Then
$$
\alpha_\gamma(s)=\int_0^s\kappa(t)dt\quad\text{and}\\
\alpha_C(s)=\int_0^skdt=ks.
$$
Because $k<\kappa(s)$, we have $\alpha_C(s)<\alpha_\gamma(s)$ for all $s>0$.

It suffices to prove that there are no arc lengths $s_\gamma$ and $s_C$ so that $\gamma(s_\gamma)\in C$ and $\alpha_\gamma(s_\gamma)\leq\pi$.
(If the turning angle of an intersection point is more than $\pi$, one can just go along the curves in the opposite direction.)
We can also exclude the case $\alpha_\gamma(s_\gamma)=\pi$, since that would correspond to $\gamma$ meeting $C$ again only at the antipodal point of $x$, which is only possible by a tangential touch.
Impossibility of tangential intersections was observed in the question.

Let $L$ be the tangent of $C$ and $\gamma$ at $x$.
The orthogonal projection of the point $\gamma(s)$ to $L$ is at (signed) distance $\int_0^s\cos\alpha_\gamma(t)dt$ from $x$.
A similar formula holds for points of points on $C$.
At the hypothetical intersection point we have
$$
\int_0^{s_\gamma}\cos\alpha_\gamma(t)dt
=
\int_0^{s_C}\cos\alpha_C(t)dt.
$$
It follows from the inequality $\alpha_C<\alpha_\gamma$ obtained above that $s_C<s_\gamma$.

It then remains to show that the arc $A=\gamma([0,s_\gamma])$ is shorter than the corresponding arc $B$ of the circle $C$ to obtain a contradiction.
The arc $A$ lies between $B$ and the line segment $S$ joining $\gamma(0)$ and $\gamma(s_\gamma)$.
Therefore $A$ is shorter than $B$.

Differential geometry of curves and surfaces. A concise guideyou'll find the ideas you need to prove the statement. $\endgroup$ – Liviu Nicolaescu Jun 3 '16 at 10:31