General Katz-Sarnak heuristics suggest that the analogue of the minimalist conjecture should still be true. Let me sketch the reason why from two perspectives - the function field model, where we can establish a version of the conjecture, and the Sarnak-Shin-Templier conjectures on families of automorphic forms.

First note that it is not obvious what you mean by a random nice curve of genus $g$, because there is not always a clear height function on the moduli space of smooth curves of genus $g$. Moreover it is not always rational, so we do not always know how to count curves of bounded height!

It's safer to choose some parameterized family of curves, like hyperelliptic curves or plane curves, and ask for the distribution in that. However, my answers will not depend too much on the family.

In both cases, we will assume the truth of BSD and show that 50% of abelian varieties have analytic rank 0 and 50% have analytic rank 1 (under some heuristic.)

In a function field, an abelian variety $A$ over $\mathbb F_q(T)$ has analytic rank equal to the dimension of the Frobenius-invariant subspace of $H^1(\mathbb P^1_{\overline{\mathbb F_q}}, T_\ell(A))$ where $T_\ell(A)$ is the $\ell$-adic Tate module of $A$. Our family of curves will be parameterized by a space like $Maps(\mathbb P^1, \mathbb P^n)$ where $\mathbb P^n$ parameterizes our family of curves over $\mathbb F_q$. This cohomology group $H^1(\mathbb P^1_{\overline{\mathbb F_q}}, T_\ell(A))$ forms a sheaf on this mapping space and we may compute its monodromy group.

If the monodromy group on a component is the full orthogonal group, the minimalist conjecture is true in the $q \to \infty$ limit for that component, because the distribution of Frobenius is as a random element in the monodromy group, and a random orthogonal matrix has a $0$-dimensional invariant subspace with probability $1/2$ and a $1$-dimensional invariant subspace with probability $1/2$. This argument is explained in Katz's book Twisted L-Functions and Monodromy.

For an explicit family it's not too hard to show that the monodromy group is orthogonal using the moment method. To show the monodromy group contains the special orthogonal group, it's sufficient to show that the first few moments agree with the moments of the orthogonal one. This can be done for sufficiently large degree maps from $\mathbb P^1$ to $\mathbb P^n$ using an independence argument, as long as $H^{2n-1}(\mathbb P^n_{\overline{\mathbb F_q}}, T_\ell(A))$ vanishes, which can be checked for the universal families of hyperelliptic and complete intersection curves by another independence argument. This sort of argument is explained in Katz's book Moments, Monodromy and Perversity.

In the note "Families of L-Functions and their Symmetry", Sarnak, Shin, and Templier conjecture equidistribution results for the L-functions of certain families of automorphic forms. Given any family of curves parameterized by an open subset of $\mathbb A^n$ (e.g. hyperelliptic, plane curves, complete intersection) the (conjectural) automorphic forms associated to their first etale cohomology groups are clearly a "geometric family" by the definition of the authors. In any of these families the local factors will be equidistributed in $SP_{2g}$, because a hyperelliptic/plane/complete intersection curve over a finite field has characteristic polynomial of Frobenius equidistributed in $SP_{2g}$.

Furthermore in their notation the rank of the family will be $0$ - this is actually the same $H^{2n-1}(\mathbb P^n_{\overline{\mathbb F_q}}, T_\ell(A))$ vanishing condition as before and can be checked in the same way.

Also, the $\epsilon$ factors should be equidistributed between $+1$ and $-1$ - I don't immediately see a reason for this but I don't see a reason why not.

Hence assuming Conjecture 2, the distribution of the low-lying zeroes follow the $SO_{even}(\infty)$ law half the time and the $SO_{odd}(\infty)$ law half the time. The first one has average analytic rank $0$ and the second has average analytic rank $1$.