This is just a partial answer to restrict the search.

**Condition** $NC_q$ Let us say that $Z$ satisfies condition $NC_q$ if for any $z_1,\dots,z_q\in Z$ there exists $D=D_{z_1,\dots,z_q}\subset Z$, a codimension one subscheme of $Z$ such that $z_i\not\in D$, for any $i=1,\dots,q$.

**Example** If $Z$ is projective, then it saisfies $NC_q$ for any $q\in\mathbb N$.

**Claim**
Suppose that $Z$ is a scheme that satisfies $NC_q$ and let $f:V\to Z$ be a dominant morphism. Then $V$ satisfies $NC_q$.

**Proof** Apply the condition for $f(v_1),\dots,f(v_q)$ and take $f^{-1}(D)\subset V$.

**Corollary** If $V$ is an arbitrary threefold that admits a dominant morphism to a projective variety of positive dimension, then it satisfies $NC_2$. In particular, Hironaka's example of a non-singular non-projective complete threefold does not provide a counter-example.

There are non-projective schemes with trivial Picard group and hence without non-trivial line bundles. Obviously these cannot admit a non-trivial morphism to a projective space (variety).

On the other hand, such a scheme has to be singular: Take an arbitrary dense affine subset and take an effective divisor there. The closure of this in the original scheme will give a non-trivial divisor which would then have to be a non-Cartier divisor, as otherwise the Picard group could not be trivial. But then the ambient scheme has to be singular.

Alternatively one could argue, using Chow's lemma that there are non-trivial divisors on any complete variety and hence the Picard group of a complete non-singular variety cannot be trivial.

So, an example of a complete non-singular threefold failing $NC_q$ would have the property that every linear system has base points. I don't know if that exists.

For a singular example you could do the following (this is motivated by Jason's comment): Take a projective threefold whose Picard group is $\mathbb Z$ and hence every effective divisor is ample on it. Contract some curves to get something non-projective. Now take any surface on the target. The pre-image of that is an effective divisor on the original threefold which then has to be ample, and hence it has to intersect the contracted curves. Therefore the chosen surface has to contain the image(s) of these curves, so if you choose your $P,Q$ among these image points, then you get a counterexample.