There is an old example due to Kochen and Specker of sort-of this point, which later was replaced by a much better example due to Bell. (Actually, the Kochen-Specker construction is related to an earlier, stronger result of Andrew Gleason.) Bell's example, after some tidying up, was actually tested in a physical experiment by Alain Aspect, and by many other people since. In the Kochen-Specker thought experiment, finite additivity does not directly fail. Rather, you can exhibit a set of Booleans which cannot generate a traditional Boolean algebra that is consistent with the quantum interpretation. In other words, you cannot have "underlying determinism" that is consistent with natural embeddings of classical probability into quantum probability.
Consider the quantum probability space $M_3 = M_3(\mathbb{C})$ of $3 \times 3$ matrices. For each line $L \in \mathbb{C}^3$, there is a corresponding Boolean event given as an operator by the orthonormal projection onto $L$. When Boolean events commute, they generate a commutative algebra, and thus a classical probability space, and one has traditional additivity. So if you have an orthonormal frame of lines, you get a copy of the Boolean algebra of events on three points. However, these Boolean algebras are inconsistent, even just using lines in $\mathbb{R}^3$. There exists a collection of 31 lines in $\mathbb{R}^3$ that do not admit a boolean function $f$ such that exactly one of $f(L_1)$, $f(L_2)$, and $f(L_3)$ is true for every orthogonal frame $\{L_1,L_2,L_3\}$. (Following Gleason, Kochen and Specker found 117 lines, but this was simplified to 31 by Conway and Kochen.) The set of lines is complicated, but there is a simpler construction of Peres in $\mathbb{R}^4$, which then establishes the contradiction for $M_4$. Namely, you should take the 12 diagonals of the regular 24-cell and the 12 diagonals of the dual 24-cell. Then it is an exercise to check that these 24 lines have the same inconsistency with respect to orthonormal frames.
At another level, the contradiction that you seek is not possible without an extra structure on classical or quantum probability. Classical probability makes a category whose objects are Boolean algebras or $L^\infty$ algebras, and whose morphisms are, say, stochastic maps. (Actually you want contravariant stochastic maps. An map from algebra $A$ to algebra $B$ yields a map of states from $B$ to $A$, and stochastic maps are usually defined on states.) Quantum probability does also, where the objects are von Neumann algebras and the morphisms are quantum stochastic maps, by definition completely positive, normal, unital maps. However, any such category embeds in the classical probability category or even the category Set. This formal result has been considered important by people who want to make quantum probability look classical. (E.g., it is essentially Bohm's point.) It means that if you have a single quantum system, you can always build a countability additive classical probability model "behind" it.
However, classical probability is also a tensor category to model joint systems, and so is quantum probability. Bell's theorem is that there does not exist a tensor embedding (speaking loosely; I'd have to think about how to rigorously define what it excludes) of quantum probability into classical probability. In physics terminology, a non-tensor-preserving construction such as that of Bohm is called "non-local".
Here is a description of an optimized form of Bell's construction called the CHSH inequality. Let $a_1, a_2 \in A$ and $b_1, b_2 \in B$ be four Booleans or, more conveniently, four $\pm 1$-valued Bernoulli random variables. Then their correlation matrix $E[a_j b_k]$ in $A \otimes B$ satisfies this elementary inequality in classical probability:
$$E[a_1 b_1] + E[a_1 b_2] + E[a_2 b_1] - E[a_2 b_2] \le 2.$$
To prove this inequality, consider that each term is $\pm 1$ if you plug in four values of the variables; if the first three term are 1 then all four are equal and the last term is $-1$. (You can easily make them unbiased so that $E[a_j b_k]$ really is the correlation matrix.) However, if you take $A \cong M_2$ (the qubit system) and $B \cong M_2$, then it is easy to find four such random variables and a state on $A \otimes B$ such that instead,
$$E[a_1 b_1] + E[a_1 b_2] + E[a_2 b_1] - E[a_2 b_2] = 2\sqrt{2}.$$
(This is the maximum possible value in quantum probability.) This is a classical impossible set of correlations. You can repeatedly interrogate system $A$ by randomly measuring either $a_1$ or $a_2$, and the same for $B$. The correlation makes you think that $A$ and $B$ are in communication with each other, even when it cannot be true. This is what was actually demonstrated in Aspect's experiment. His experiment directly violated the CHSH inequality, in a protocol in which there wasn't enough time for light to travel from $A$ to $B$.
Here are a few more comments about the violation demonstrated in Aspect-type experiments. What is actually measured is
$$E = E[(-1)^{(j+1)(k+1)} a_j b_k],$$
where $j$ and $k$ are also random variables taking values in $\{1,2\}$. If $A$ and $B$ (or Alice and Bob) could communicate, then they could easily make $E > \frac12$, or even as close to 1 as they please, because they would both know the pair $(j,k)$ and they could pick a favorable pair of answers. But in the experiment, they are separated, and Alice is only told $j$ and Bob is only told $k$. In this case, they must separately provide $a_j$ and $b_k$, and the CHSH inequality applies if $A$ and $B$ are classical.
This is a somewhat formal and blasé summary of the illusion of telepathy in quantum probability that was first constructed by Bell. I have a livelier description of the same formulas in a colloquium talk that I gave at Berkeley.
