I thought about this question some yesterday. As I was saying in the related post, von Neumann algebras are a non-commutative or quantum generalization of measurable spaces,
$C^*$ algebras are a non-commutative/quantum generalization of compact Hausdorff spaces, and both generalizations are contravariant. Your question has a covariant spirit, which is a bit misaligned but not an essential point for this particular question.
The more germane issue is that the interval is a topological space in Peter Freyd's construction, while the hyperfinite factor, like any von Neumann algebra, behaves as a measurable space. It is true that the hyperfinite factor is a non-commutative analogue of an interval, and you can construct it as the closure of a certain class of operators on $L^2([0,1])$. But Freyd's construction does not work for measurable maps, only continuously. In fact, as a measurable space, the unit interval doesn't have endpoints. It has points, sort-of, but not in any useful way. (To be precise, the measurable model of the interval here is the class of measures on it which are Lebesgue absolutely continuous, not all Borel measures.)
Maybe there is a
$C^*$-algebra with a Freyd-type property, and which generates the hyperfinite factor. You would have to decide whether its "endpoints" are a classical bit or a qubit. You could try to work in the category of
$C^*$-algebra homomorphisms, which tends to be short of morphisms from non-commutative objects. Or you could try to work in the category of completely positive maps on
$C^*$-algebras, which has plenty of morphisms but also has other complications. Certainly you would want to reverse arrows in passing from topology to $C^*$-algebras. I don't know a whole lot about making
$C^*$-algebras; I couldn't come up with anything.