Are there irreducible tensors of half integral degree in quantum mechanics?

Yes, these "tensors" are called "spinors" for $k=1/2$ or perhaps "spintensors" for $k\in{\mathbb Z}+1/2$ but they satisfy the very same algebra you wrote. It is somewhat misleading to ask about "a simple example". There's just one example – the example or the solution – for each value of $k$, up to the choice of a basis and up to the freedom to pick a greater degeneracy. And you have actually written it down.

The collection of $k=1/2$ operators is $2k+1=2$-dimensional and it's known as the spinor. The allowed values are $q=+1/2$ and $q=-1/2$ and your formulae actually describe the exact form of all the commutators for $k=1/2$ as well as any other allowed value of $k$ (integer or half-integer-valued).

Spinors are essential to describe elementary fermions, including electrons, neutrinos, and quarks (and even protons, neutrons, and all other composite spin-1/2 particles). These may be viewed as "many examples" but the maths how the $\vec J$ generators commute with the spinorial objects is exactly the same in all cases.

The reason why one can construct tensors with a "half-integer number of indices" is that the group $SO(3)$ of rotations, generated by the angular momentum, is isomorphic to $SU(2)/{\mathbb Z}_2$. The quotient by ${\mathbb Z}_2$ means that one only allows integer spins. However, if one allows representations of the whole $SU(2)\sim Spin(3)$, one also allows representations (and tensor operators) that change their sign after rotations by 360 degrees and they correspond to $k$ which differs by $1/2$ from an integer.

1. Yes, I assume that OP is already familiar with the irreducible finite-dimensional representations $V_j$ of the Lie algebra $so(3)\cong su(2)$; that they are classified by a non-negative half-integer $j\in \frac{1}{2}\mathbb{N}_{0}$; and that the eigenstates $|j,m\rangle \in V_j$ for $\vec{J}^2$ and $J_z$ satisfy $$\begin{align} \vec{J}^2 |j,m\rangle ~=~& \hbar^2 j(j+1)|j,m\rangle , \cr J_z |j,m\rangle ~=~& \hbar m|j,m\rangle ,\cr J_{\pm}|j,m\rangle ~=~& \hbar \sqrt{ (j \mp m)(j \pm m + 1)}|j,m\pm 1\rangle , \end{align}$$ where $-j\leq m\leq j$ such that $j-m\in\mathbb{Z}$.

2. Instead, it seems that OP is really asking about the tensor operators $T_q{}^{(k)}$ described in eq. (7.112) of the book Quantum mechanics: a modern development by Leslie E. Ballentine, $$\begin{align}[J_{\pm}, T_q{}^{(k)}] ~=~& \hbar \sqrt{ (k \mp q)(k \pm q + 1)} T_{q \pm 1}{}^{(k)},\cr [J_z, T_q^{\;\;(k)}] ~=~& \hbar q T_{q}{}^{(k)}.\end{align} \tag{7.112} $$

3. A simple example of such operators $$\begin{align}T_q{}^{(k)}:&~V~\to~ V, \cr V~=~& \oplus_{j\in \frac{1}{2}\mathbb{N}_{0}} V_j,\end{align}$$ can be build as $$T_q{}^{(k)}~:=~ | k,q\rangle \otimes \langle 0,0|,$$ where $k\in \frac{1}{2}\mathbb{N}_{0}$ is a non-negative half-integer, and $-k\leq q\leq k$ such that $k-q\in\mathbb{Z}$. It is straightforward to check that this example satisfies the wanted commutation relations (7.112).

4. Note that any set of $2k+1$ operators $T_q{}^{(k)}:V_0\to V_k$ with $-k\leq q\leq k$ that satisfy eq. (7.112) would have to be proportional to $| k,q\rangle \otimes \langle 0,0|$ with the same proportionality factor independent of $q$, due to Wigner–Eckart theorem.