If we define $\mathcal{T}=i\sigma_y K$ where $K$ is complex conjugation, i.e.
$\mathcal{T}\psi_\uparrow \mathcal{T}^{-1}=\psi_\downarrow, \mathcal{T}\psi_\downarrow \mathcal{T}^{-1}=-\psi_\uparrow$,
Then naively a term like $\Delta e^{i\Phi}\psi_{\uparrow}^\dagger \psi_{\downarrow}^\dagger$ is not invariant under $\mathcal{T}$. This is basically the problem you encountered, phrased a little differently. However, it does not mean that the system actually breaks $T$, as all physical observables will be invariant under $\mathcal{T}$. The resolution is in the definition of the transformation of $\psi$ under $\mathcal{T}$. Let us modify the definition to be
$\mathcal{T}\psi_\uparrow \mathcal{T}^{-1}=\psi_\downarrow e^{-i\Phi}, \mathcal{T}\psi_\downarrow \mathcal{T}^{-1}=-\psi_\uparrow e^{-i\Phi}$.
This phase is not observable (it is essentially redefining the phases of the basis states of the second-quantized Fock space, which has no consequences on physical observables), so we are free to do so. Notice that the important algebraic relation $\mathcal{T}^2=-1$ (for the classification of TI/TSC, etc.) is not affected. Then the pairing term is invariant.
Of course, this only works when $\Phi$ does not depend on positions. Otherwise (e.g. when there is a vortex) one can not get rid of the phase, since $\nabla\Phi$ is an observable, the supercurrent.
This is essentially an issue of first quantization vs. second quantization. In second quantization we define the action of the time-reversal operator on the fermionic creation and annihilation operators to be
\begin{equation}
\begin{split}
\hat{\mathcal{T}}\hat{c}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}_{i, \downarrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \uparrow}\hat{\mathcal{T}}^{-1} = \hat{c}^{\dagger}_{i, \downarrow} \\
\hat{\mathcal{T}}\hat{c}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}_{i, \uparrow}, \quad \hat{\mathcal{T}}\hat{c}^{\dagger}_{i, \downarrow}\hat{\mathcal{T}}^{-1} = -\hat{c}^{\dagger}_{i, \uparrow}.
\end{split}
\end{equation}
where $\hat{c}^{\dagger}_{i, \sigma}/\hat{c}_{i, \sigma}$ are the fermionic creation and annihilation operators acting on site $i$ and spin state $\sigma=\uparrow/\downarrow$.
In addition $\hat{\mathcal{T}}$ is an antiunitary operator, therefore $\hat{\mathcal{T}}i\hat{\mathcal{T}}^{-1} = -i$ (we can show this by considering the Heisenberg uncertainty relationship and using the relations $\hat{\mathcal{T}}\hat{x}\hat{\mathcal{T}}^{-1} = \hat{x}$ and $\hat{\mathcal{T}}\hat{p}\hat{\mathcal{T}}^{-1} = -\hat{p}$).
We can summarise the actions of $\hat{\mathcal{T}}$ on the Fock space by first converting $\hat{c}_{i, \uparrow}, \hat{c}_{i, \downarrow}, \hat{c}^{\dagger}_{i, \uparrow}, \hat{c}^{\dagger}_{i, \downarrow}$, ... to $\hat{\psi}_{1}, \hat{\psi}_{2}, \hat{\psi}_{3}, \hat{\psi}_{4}$...
The action of $\hat{\mathcal{T}}$ is then surmised as $\hat{\mathcal{T}}\hat{\psi}_{A}\hat{\mathcal{T}}^{-1} = \sum_{B} U_{A, B}\hat{\psi}_{B}$ where $A, B$ are indices labelling the site and other relevant quantum numbers such as spin and $U$ is some unitary matrix. See Kitaev's paper for more details https://arxiv.org/abs/0901.2686. It should be emphasised that, so far, we have defined the action of time-reversal on the Fock space in the formalism of second quantization.
To move to the first quantized picture, we write the second quantized Hamiltonian, $\hat{H}$ in terms of the single-particle Hamiltonian as
\begin{equation}
\hat{H} = \sum_{A, B} \hat{\psi}_{A}^{\dagger}H_{A, B}\hat{\psi}_{B}
\end{equation}
where the operators $\hat{\psi}^{\dagger}_{A}/\hat{\psi}_{A}$ satisfy the usual anticommutation relations and $H$ is the first quantized Hamiltonian (basically just an $N\times N$ matrix). If the second quantized Hamiltonian possesses time reversal symmetry then $\hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} = \hat{H}$. Using $\hat{\mathcal{T}}\hat{\psi}_{A}\hat{\mathcal{T}}^{-1} = \sum_{B} U_{A, B}\hat{\psi}_{B}$ we find
\begin{equation}
\begin{split}
\hat{\mathcal{T}}\hat{H}\hat{\mathcal{T}}^{-1} &= \sum_{A, B} \hat{\mathcal{T}} \hat{\psi}^{\dagger}_{A} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}} H_{A, B} \hat{\mathcal{T}}^{-1} \hat{\mathcal{T}} \hat{\psi}_{B} \hat{\mathcal{T}}^{-1} \\
&= \sum_{A, B}\sum_{C, D}U^{*}_{A, C} \hat{\psi}^{\dagger}_{C} H^{*}_{A, B} U_{B, D} \hat{\psi}_{D} \\
&= \sum_{C, D} \hat{\psi}^{\dagger}_{C} H_{C, D} \hat{\psi}_{D} = \hat{H}
\end{split}
\end{equation}
where $H_{C, D} = U^{*}_{A, C} H^{*}_{A, B} U_{B, D}$, i.e. $H = U^{\dagger}H^{*}U$. Here, the single particle Hamiltonian is complex conjugated since $\hat{\mathcal{T}}$ acts on the numerical parameters and reverses the sign of i. Therefore, we may define a first quantized version of $\hat{\mathcal{T}}$ which acts on the single particle space
\begin{equation}
T = \hat{\mathcal{T}}_{first quantized}
\end{equation}
We may then rewrite the action of time reversal on the first quantized Hamiltonian as
\begin{equation}
THT^{-1} = H \quad \text{where} \quad T = UK
\end{equation}
In summary, complex conjugation acts on numbers rather than operators.
For more information on the time reversal operator and its use in classifying topological phases of matter see these excellent review papers:
https://arxiv.org/abs/0912.2157
https://arxiv.org/abs/1512.08882
https://doi.org/10.1103/RevModPhys.88.035005
Best Answer
The transformation under time reversal of the forms in electrodynamics is subtle because the gauge field 1-form $A = A_\mu \mathrm{d}x^\mu$ and the field strength $F = F_{\mu\nu}\mathrm{d}x^\mu\wedge\mathrm{d}x^\nu$ are not the correct physical objects to transform.
This may be seen by observing that the Maxwell equations are $\mathrm{d}F = 0$ and $\mathrm{d}\star F = \star J$, but the former is just a Bianchi identity following from $\mathrm{d}^2 = 0$. The actual equation of motion for the gauge theory is given in terms of the Hodge duals $\star F$ and $\star J$, and it are thus the Hodge duals whose transformation behaviour dictates the transformation behaviour under time reversal.
In the field strength tensor, we have the terms $E_i \mathrm{d}t\wedge\mathrm{d}x^i$ and $B_i \epsilon^{ijk}\mathrm{d}x^j\wedge\mathrm{d}x^k$ and from this one would indeed conclude that it is the electric field that changes sign under time reversal. However, inspecting the Hodge dual that occurs in the equation of motion, we find the opposite behaviour since the star of the terms with $\mathrm{d}t$ contains no $\mathrm{d}t$ terms anymore and vice versa.
This highlights a general and important fact: The Hodge star does not commute with coordinate transformations that change the handedness of the underlying coordinate system, since its definition crucially relies on the ordering and handedness of the vectors in the system. Therefore, as soon as we consider transformations whose determinant is negative (since that is the abstract sign of changing handedness), care must be taken for all geometric objects whether the correct physical interpretation is to have the transformation act on them or their duals.