Let me show that in the case you blow up $\mathbb P^2$ in one point an throw away preimages of three lines trough the point you get something with non-abelian fundamental group. The blow up of $\mathbb P^2$ at one point is a $\mathbb P^1$ bundle over $\mathbb P^1$, so when we throw away $3$ fibers we get a fibration $\mathbb P^1\to S\to \ C$, where $C$ is $\mathbb P^1$ punctured in $3$ points. The long exact sequence of fundamental groups reads as follows:

$...\to \pi_1(\mathbb P^1)\to \pi_1(S)\to \pi_1(C)\to \pi_2 \mathbb (\mathbb P^1)\to ...$

$...\to 0 \to \pi_1(S)\to \pi_1(C)\to \mathbb Z \to ...$

In particular, we see that $\pi_1(S)$ injects intp the free group $F_2$ on $2$ generators and it contains the commutator group of $F_2$ (since the commutator of $F_2$ goes to zero under $F_2\to \mathbb Z$). Hence $\pi_1(S)$ is not commutative.

I can not say anything about the general situation...

Corrected according to the comment of Scott.

Let me show that in the case you blow up $\mathbb P^2$ in one point an throw away preimages of three lines trough the point you get something with non-abelian fundamental group. The blow up of $\mathbb P^2$ at one point is a $\mathbb P^1$ bundle over $\mathbb P^1$, so when we throw away $3$ fibers we get a fibration $\mathbb P^1\to S\to \ C$, where $C$ is $\mathbb P^1$ punctured in $3$ points. The long exact sequence of homothopy groups reads as follows:

$...\to \pi_1(\mathbb P^1)\to \pi_1(S)\to \pi_1(C)\to \pi_0 \mathbb (\mathbb P^1)\to ...$

$...\to 0 \to \pi_1(S)\to \pi_1(C)\to 0 $

So $\pi_1(S)=F_2$, where $F_2$ is a free group on two generators.

I can not say anything about the general situation...