Noncommutative torus as a von Neumann algebra
No. It's irreducible. The element $U$ generates the maximal abelian subalgebra $L^\infty({\mathbb T})$ and hence one computes the commutant: $$\{U,V\}'=\{U\}'\cap\{V\}'=L^\infty({\mathbb T})\cap\{V\}'={\mathbb C}1.$$ By the way, the invariant subspace problem for the Bishop operator $f(x)\mapsto xf(x+\theta)$ is still open in full generality. https://mathscinet.ams.org/mathscinet-getitem?mr=353015
To support Ruy's answer: in my opinion the most natural representation of the quantum torus C*-algebra is the GNS representation coming from its tracial state. This can be explicitly described as the action on $l^2(\mathbb{Z}^2)$ given by $$Ue_{m,n} = e^{-i\hbar n/2}e_{m+1,n}$$ and $$Ve_{m,n} = e^{i\hbar m/2}e_{m,n+1}.$$ The von Neumann algebra they generate is indeed a $II_1$ factor.
I would even say this is the "quantum torus von Neumann algebra". There's more in Section 6.6 of my book Mathematical Quantization.
As I was reading Yemon Choi's comment above it occurred to me that the situation of the crossed product $C(S^1)\times_\theta\mathbb{Z}$ is in fact a bit peculiar since the most standard representation of $C(S^1)$ one usually thinks of, namely as multiplication operators on $L^2(S^1)$, already comes equipped with a unitary representation of $\mathbb{Z}$ implementing the action by rotation.
This is not always the case for a general crossed product $A\times\mathbb{Z}$, so one usually starts with one's favorite representation of $A$ on some Hilbert space $H$ and builds the "regular representation" of the crossed product on the Hilbert space $H\otimes \ell^2(\mathbb{Z})$.
Even though that was not the representation the OP had in mind it is interesting to observe that, if the irrational rotation C*-algebra is completed in this other representation, one does indeed get a type $II_1$ factor, partly because the standard trace is a vector state in this representation and hence duly extends to a normal state on the weak closure.
PS: It was my original intention to reply to a comment by Yemon Choi, but I could not fit all of this within the 600 charactes size limitation. I therefore hope to be excused for shamelessly attempting to sidestep the rules and I am ready to delete this post should anyone complain!