I think the answer should be 'no'.
Because when we introduce the antiunitary time-reversal(TR) opeartor $T$ for spin-system, it should satisfy $T\mathbf{S}_iT^{-1}=-\mathbf{S}_i$ since angular-momentum should be sign reversed under TR(due to the classical correspondence). Thus, spin-spin interactions like $\mathbf{S}_i\cdot\mathbf{S}_j$ are invariant under TR.
The TR operator $T$ for the $N$-spin-$1/2$ system has a form $T=(-i)^N\sigma_1^y\sigma_2^y...\sigma_N^yK$, where $K$ is the conjugation operator. You can easily check that $T$ is antiunitary and satisfy $T\mathbf{S}_iT^{-1}=-\mathbf{S}_i$. Furthermore, $T^2=(-1)^N$, so for odd-number spin system(including single spin case), if the Hamiltonian has TR symmetry, we will arrive at the well-known Kramers theorem.
I don't know the article you refer to, but I believe the Hamiltonian you discuss should get a $\pi$-phase shift after one turn around a (2D) lattice cell. So I guess it should read $H=F^{\dagger}\cdot H_{\pi}\cdot F$ with
$$H_{\pi}=t\left(\begin{array}{cccc}
0 & e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4}\\
e^{-\mathbf{i}\pi/4} & 0 & e^{\mathbf{i}\pi/4} & 0\\
0 & e^{-\mathbf{i}\pi/4} & 0 & e^{\mathbf{i}\pi/4}\\
e^{\mathbf{i}\pi/4} & 0 & e^{-\mathbf{i}\pi/4} & 0
\end{array}\right)$$
and $F^{\dagger}=\left(\begin{array}{cccc}
f_{1}^{\dagger} & f_{2}^{\dagger} & f_{3}^{\dagger} & f_{4}^{\dagger}\end{array}\right)$. Then, one has
$$H_{\pi}=\dfrac{t}{\sqrt{2}}\left[\left(1+\tau_{x}\right)\otimes\eta_{x}-\left(1-\tau_{x}\right)\otimes\eta_{y}\right]$$
where the $\eta$ and $\tau$ are the usual Pauli matrices.
Time reversal symmetry operator -- when it exists -- is defined as an anti-unitary operator which commutes with the Hamiltonian. Such an operator can be defined as $T=\mathscr{K}\tau_{z}\otimes\mathbf{i}\eta_{y}$ and thus $H$ is time reversal symmetric. $\mathscr{K}$ is the anti-unitary operator $\mathscr{K}\left[\mathbf{i}\right]=-\mathbf{i}$ and thus $\mathscr{K}\left[\eta_{y}\right]=-\eta_{y}$. One verifies that $\left[H_{\pi},T\right]=0$ as it must.
Please tell me if I started with the wrong Hamiltonian.
A few words about the definition (as follow from the comment below): The time-reversal operator is defined as I did, i.e. one applies it to the Hamiltonian $H_{\pi}$, (call it the Hamiltonian density if you wish, since in my way of writing $H=F^{\dagger}\cdot H_{\pi}\cdot F$, the dots should include summation(s) over phase-space-time [delete as appropriate]). You could prefer to define the action of an operator as transforming the operators (or the wave-function). But you should not use both definitions at the same time. It is clear that you can not do both, since otherwise you transform $H=F^{\dagger}\cdot H_{\pi}\cdot F \rightarrow F^{\dagger}\cdot U^{\dagger}\cdot \left(U \cdot H_{\pi} \cdot U^{\dagger}\right) \cdot U\cdot F = H$ trivially, whatever (anti-)unitary transformation $U$ you choose. It is clear that what your are looking for is something like $H=F^{\dagger}\cdot H_{\pi}\cdot F \rightarrow F^{\dagger}\cdot U^{\dagger}\cdot H_{\pi} \cdot U\cdot F \sim H$ and you see what I just said: apply the transformation to the Hamiltonian (density) or to the fields, but not both. In condensed matter we usually choose the convention I gave to you: we transform the Hamiltonian. One of the reasons is that the operators (especially the fermionic creation/annihilation ones) are seen as encoding the statistics of the fields, whereas the Hamiltonian encodes the dynamics, and it is simple imagination to change the dynamics.
Best Answer
Before you start your symmetry analysis, you have to define which system you consider. In this case, do you only consider the spin $1/2$-particle, or do you consider both the particle and the magnetic field along with its source?
If you consider both the particle and the magnetic field along with its source as your system, then yes, the system does have a time-reversal symmetry acting on all components.
If you only consider the particle as your system, then the external magnetic field must be kept invariant under time-reversal symmetry. In that case, the system is not time reversal invariant.
For spin-orbit coupling, the only reasonable choice is to choose the particle as your system. Then both momentum and spin change sign under time reversal.