Fix a system of homogeneous coordinates: $${\bf P}^1_{\mathbb Z}=\mathrm{Proj\ } \mathbb Z[x,y].$$
Every rational point $b$ of ${\bf P}^1_{\mathbb Q}$ can be written in homogeneous coordinates as $(u : v)$ with coprime integers $u, v\in \mathbb Z$. The vector $(u, v)$ is unique up to signs. The closure $P$ of $b$ has integral homogeneous coordinates $(u:v)$ and the image of $P$ in ${\bf P}^1_{\mathbb F_p}$ has homogeneous coordinates $(\bar{u}: \bar{v})$ with obvious meaning of $\bar{u}$ and $\bar{v}$.
Now write $b_1=(u_1:v_1)$, $b_2=(u_2:v_2)$ as above. Then $P_1$ meets $P_2$ above a prime number $p$ if and only if $(\bar{u}_1:\bar{v}_1)=(\bar{u}_2:\bar{v}_2)$ in ${\bf P}^1$ over ${\mathbb F}_p$. Or equivalently if $u_1v_2-v_1u_2 \equiv 0 \mod p$. Therefore $P_1\cap P_2=\emptyset$ if and only if $u_1v_2-v_1u_2=\pm 1$.
A more abstract characterization is, as you suggest, there exists an automorphism of ${\bf P}^1_{\mathbb Z}$ which maps $P_1$ to $(0:1)$ and $P_2$ to $(1:0)$. This is obviously sufficient. The above descriptions shows that if $P_1$ doesn't meet $P_2$, then the automorphism $(x:y)\mapsto (v_1x-u_1y : v_2x-u_2y)$ of ${\bf P}^1_{\mathbb Z}$ maps $b_1$ to $(0:1)$ and $b_2$ to $(1:0)$.
Note that one can't expect a characterization uniquely in terms of the generic fiber because the condition on $P_1, P_2$ depends on a choice of the model ${\bf P}^1_{\mathbb Z}$. For example, $\mathrm{Proj\ } \mathbb Z[2x,y]$ is isomorphic to ${\bf P}^1_{\mathbb Z}$, the points $(1:1), (0:1)$ (in the coordinates $x,y$) meets, in this new model, above $p=2$, but they don't meet in $\mathrm{Proj\ } \mathbb Z[x,y]$.