Consider $B$ a finite-type integral quasi-projective scheme over $\mathbb{Z}$ such that $B(\mathbb{Z})$ infinite. (If you like, take $B$ to be the affine line). Let $X \to B$ be a generically smooth proper family of curves of genus $g > 1$. Assume the $b \in B(\mathbb{Z})$ for which $X_b(\mathbb{Q}) = \emptyset$ are Zariski-dense in $B$. 
Must then the proportion of members $X_b$, over $b \in B(\mathbb{Z})$, for which $X_b(\mathbb{Q}) \neq \emptyset$, be equal to $0$? (Variant: the same question with $B(\mathbb{Z})$ replaced by $B(\mathbb{Q})$.)

*An explicit variant (although not quite a special case as it stands)*. Let $f \in \mathbb{Z}[x]$ be irreducible of degree $> 4$. Do the integers $N$ for which $f(x) = Ny^2$ has a rational solution, have density zero? Variant: let $N$ range over the primes, or over the squarefrees.

In the above setup, we may include this example by assuming more generally that $X_{\mathbb{Q}} \to B_{\mathbb{Q}}$ admits exactly $m$ sections defined over $\mathbb{Q}$, and ask whether the density of $b \in B(\mathbb{Z})$ with $|X_b(\mathbb{Q})| > m$ is zero.

*A comment.* This, of course, is easily seen to fail for families of genus zero. For example, if $\deg{f} = 2$ above, then for half the primes $N$ (those that split in the quadratic field $\mathbb{Q}[x]/(f(x))$) the equation $f(x) = Ny^2$ has a solution. Various heuristics, and some standard conjectures, imply that it should also fail for genus one families. I would be interested in seeing an example in genus one for which it can be *proved* that a positive proportion (yet not almost all, for the Zariski topology) of the members have rational points -- or is this out of reach?