Let H be an infinite dimensional Hilbert space. Then there exist non-normal states on B(H) in ZFC (i.e. states that are not represented by a density operator).
Is this also true in the Solovay model ?
I don't think so but I couldn't find a reference.
The question is essentially already answered in the comments and answers, but I thought it might be valuable to combine these observations into something definitive.
Solovay's model is not unique. The model produced by Solovay's construction depends on the choice of ground model, and on the choice of generic filter. We can always arrange that the ground model satisfies the axiom of constructibility, and when this is the case, it is possible to establish many familiar theorems in the resulting Solovay model by appealing to absoluteness. This is discussed in my paper V*-algebras. There, I work in the Chang model, but as I explain in the first paragraph of section 5, all the results that I state in the Chang model are also true in a Solovay model of the kind I just described, except for the axiom of determinacy.
In such a Solovay model, all states on $\mathcal B(\mathcal H)$ are normal, whenever $\mathcal H$ is separable (remark 8.41). If $\mathcal H = \ell^2([0,1])$, then the state $\lambda\colon x\mapsto \int_0^1\langle e_t| x e_t\rangle dt$ is not normal, as jjcale suggested, because the projections onto the subspaces $\mathbb C e_t$ sum to the identity, but $\lambda$ vanishes on each of these projections. Assuming sufficient large cardinal axioms, the Chang model satisfies the axiom of determinacy, so $\omega_1$ is a measurable cardinal, which yields a non-normal state on $\mathcal B(\mathcal \ell^2(\omega_1))$, as Ashutosh suggested. This state is even less normal than $\lambda$, in the sense that $\lambda$ is normal for well-ordered sequences, but this state is obviously not.
This answers the question just for the Solovay models obtained from ground models satisfying the axiom of constructibility. I would guess that this answer is also correct for the other Solovay models.
This summarizes whatever half-baked ideas I had:
There is a non normal state on $B(H)$ for some Hilbert space $H$ of infinite dimension iff either there is a real valued measurable cardinal or there is a finitely additive probability measure on integers that vanishes on singletons.
Reference: https://math.berkeley.edu/~solovay/Preprints/Gleason_abstract.pdf
Apart from the issue with non-separable Hilbert spaces mentioned in the other answer, usually the non-normal states of a (general) von Neumann algebra $U$ are the (positive, with norm one) elements of the dual $U^*$ that are not in the predual $U_*$.
Now many algebras of interest in physics are non-separable (even if they act on separable Hilbert spaces), e.g. the algebra of the canonical commutation relations. I do not know if there is need of uncountable choice to choose elements of the dual and predual. My (very poor) logic intuition would say that you need it. Also, I am not completely sure that you can prove that each von Neumann algebra has a predual without uncountable choice.
However, if I could interpret the idea of the OP, since in the Solovay model the dual of $L^\infty$ is $L^1$ maybe this duality holds also for $\mathcal{B}(H)$ and $\mathcal{L}^1(H)$ (trace-class operators), at least when they are separable. If this is the case, no non-normal state would exist (in the sense above).
