Clearly the etale fundamental group of $\mathbb{P}^1_{\mathbb{C}} \setminus \{a_1,...,a_r\}$ doesn't depend on the $a_i$'s, because it is the profinite completion of the topological fundamental group. Does the same hold for when I replace $\mathbb{C}$ by a finite field? How about an algebraically closed field of positive characteristic?
(note that I'm talking about the full $\pi_1$ and not the prime-to-$p$ part)
It is a result of Tamagawa that for two affine curves $C_1, C_2$ over finite fields $k_1,k_2$ any continuous isomorphism $\pi_1(C_1)\rightarrow \pi_1(C_2)$ arises from an isomorphism of schemes $C_1\rightarrow C_2$. Hence, if $\pi_1( \mathbb{P}^1\setminus\{a_1,\ldots, a_r\})$ were independent of the choice of the $a_i$, then the isomorphism class of the schemes $\mathbb{P}^1\setminus\{a_1,\ldots, a_r\}$ would be independent of the choice of $a_1,\ldots,a_r$.
Tamagawa's result is Theorem 0.6 in this paper:
The Grothendieck conjecture for affine curves, A Tamagawa - Compositio Mathematica, 1997 http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=298922
In the case of an algebraically closed field, the answer is also that the fundamental group depends on the choice of the points that are being removed. Again by a theorem by Tamagawa: If $k$ is the algebraic closure of $\mathbb{F}_p$, and $G$ a profinite group not isomorphic to $(\hat{\mathbb{Z}}^{(p')})^2\times \mathbb{Z}_p$, then there are only finitely many $k$-isomorphism classes of smooth curves $C$ with fundamental group $G$ (the restriction on $G$ excludes ordinary elliptic curves).
This can be found in
Finiteness of isomorphism classes of curves in positive characteristic with prescribed fundamental groups, A Tamagawa - Journal of Algebraic Geometry, 2004
No --- given two triples of Q-rational points, there is an automorphism of the projective line over Q carrying one to the other.
