Let us prove that any non-simply connected surface without curves of negative self-intersection provides a counterexample. This includes Abelian surfaces and generic products of curves of higher genus.

Proof. Suppose that $X$ is a surface with isolated singularities in $\mathbb CP^3$. Then the sinuglarities of the surfaces are normal. Hence the minimal resolution of these singularities must have curves with negative self-intersections (that project to singularity). So if the resolution does not have such curves, $X$ is non-singular. It follows that $X$ has trivial fundamental group.

Correction. I see now that in order for this reasoning to work one has to guarantee that on the minimal resolution of singularities of $X$ there are no rational $-1$-curves. I wonder if this holds indeed provided the degree of $X$ is high enough. Or maybe one can "classify" surfaces in $\mathbb CP^3$ for which this statement fails. So I still believe this idea can work out...

This answer is completely rewritten. This is not an actual answer but a thought related to the question. I decided to leave it hear since it is short.

Note first that if there is a regular map from a surface $X$ to $\mathbb P^3$ whose image has only isolated singularities, then $X$ has curves with negative self-intersection. In particular, if $X$ has no such curves then its image in $\mathbb P^3$ is smooth.

Now, suppose we have a surface $X$ with isolated singularities in $\mathbb CP^3$, say of general type and consider the question:

Question. Let $X'$ be the minimal resolution of singularities on $X$. Can we say something about $X$ if $X'$ contains rational $-1$ curves?

Let us prove that any non-simply connected surface without curves of negative self-intersection provides a counterexample. This includes Abelian surfaces and generic products of curves of higher genus.

Proof. Suppose that $X$ is a surface with isolated singularities in $\mathbb CP^3$. Then the sinuglarities of the surfaces are normal. Hence the minimal resolution of these singularities must have curves with negative self-intersections (that project to singularity). So if the resolution does not have such curves, $X$ is non-singular. It follows that $X$ has trivial fundamental group.

Let us prove that any non-simply connected surface without curves of negative self-intersection provides a counterexample. This includes Abelian surfaces and generic products of curves of higher genus.

Proof. Suppose that $X$ is a surface with isolated singularities in $\mathbb CP^3$. Then the sinuglarities of the surfaces are normal. Hence the minimal resolution of these singularities must have curves with negative self-intersections (that project to singularity). So if the resolution does not have such curves, $X$ is non-singular. It follows that $X$ has trivial fundamental group.

Correction. I see now that in order for this reasoning to work one has to guarantee that on the minimal resolution of singularities of $X$ there are no rational $-1$-curves. I wonder if this holds indeed provided the degree of $X$ is high enough. Or maybe one can "classify" surfaces in $\mathbb CP^3$ for which this statement fails. So I still believe this idea can work out...