Your question is actually best reformulated in terms of horospheres and horoballs. Indeed, the set of points described in normal coordinates as $\{(x,0), x\ge 0\}$ is precisely a geodesic ray $\gamma$ in the hyperbolic plane issued from the point $p$. Then the union of the interiors of all your $C_x$ is precisely the horoball in the hyperbolic plane centered at the limit point $\gamma_\infty\in\partial \mathbf H^2$ of the ray $\gamma$ and passing through the point $p$.
Now it is more convenient (at least for me) to use the upper hlaf space model, and take $p=i$ and $\gamma$ to be the vertical geodesic ray directed upwards, so that $\gamma_\infty=\infty$. Then the corresponding horosphere (or rather the horocycle, as its dimension is 1) is just the horizontal line $y=1$, and it is clear that even if the directing vector of a geodesic issued from $p$ has a positive vertical component, then the resulting endpoint still may very well end up to be below the horosphere $y=1$. Therefore, the answer to your first question is "no".
As for your second question, by applying some elementary hyperbolic geometry one can explicitly describe all the tangent vectors at the point $p=i$ (I continue talking about the upper half space model) such that the endpoints of the corresponding geodesic are within the horoball centered at $\infty$ and passing through $p$. Let $(x,y)$ be the coordinates of the tangent vector (since I am using the upper half space model, $x$ and $y$ are swapped here compared to your original formulation), and let $\gamma$ be the geodesic issued from $p$ in the direction $(x,y)$. Then the condition is that $\gamma(d)$ is inside the horoball (i.e., its vertical coordinate is $>1$), where $d=\sqrt{x^2+y^2}$ is the length of the tangent vector. In other words, $d$ should be less than the hyperbolic distance between $p$ and the point $p'$ where $\gamma$ intersects the horosphere $\{y=1\}$. The geodesic $\gamma$ is an arc of the circle centered at $(\frac{y}x, 0)$, whence $p'=(2\frac{y}x,1)$, so that
$$
d(p,p') = arcosh \bigl( 1+ 2 (\tfrac{y}x)^2 \bigr)
$$
(e.g., see). Thus, the condition is
$$
\sqrt{x^2+y^2} > arcosh \bigl( 1+ 2 (\tfrac{y}x)^2 \bigr)
$$
or
$$
\cosh \sqrt{x^2+y^2} > \bigl( 1+ 2 (\tfrac{y}x)^2 \;.
$$
You are asking whether, for a given $x$ there is a threshold value $y_0=y_0(x)$ separating "good" and "bad" point (once again, I remind, that $x$ and $y$ are swapped compared to your original question). I am not sure this is true though. For instance, for $x=1$ the corresponding equation seems to have two roots.
