I) First of all, one should never use the Dirac bra-ket notation (in its ultimate version where an operator acts to the right on kets and to the left on bras) to consider the definition of adjointness, since the notation was designed to make the adjointness property look like a mathematical triviality, which it is not. See also this Phys.SE post.
II) OP's question(v1) about the existence of the adjoint of an antilinear operator is an interesting mathematical question, which is rarely treated in textbooks, because they usually start by assuming that operators are $\mathbb{C}$-linear.
III) Let us next recall the mathematical definition of the adjoint of a linear operator. Let there be a Hilbert space $H$ over a field $\mathbb{F}$, which in principle could be either real or complex numbers, $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$. Of course in quantum mechanics, $\mathbb{F}=\mathbb{C}$. In the complex case, we will use the standard physicist's convention that the inner product/sequilinear form $\langle \cdot | \cdot \rangle$ is conjugated $\mathbb{C}$-linear in the first entry, and $\mathbb{C}$-linear in the second entry.
Recall Riesz' representation theorem: For each continuous $\mathbb{F}$-linear functional $f: H \to \mathbb{F}$ there exists a unique vector $u\in H$ such that
$$\tag{1} f(\cdot)~=~\langle u | \cdot \rangle.$$
Let $A:H\to H$ be a continuous$^1$ $\mathbb{F}$-linear operator. Let $v\in H$ be a vector. Consider the continuous $\mathbb{F}$-linear functional
$$\tag{2} f(\cdot)~=~\langle v | A(\cdot) \rangle.$$
The value $A^{\dagger}v\in H$ of the adjoint operator $A^{\dagger}$ at the vector $v\in H$ is by definition the unique vector $u\in H$, guaranteed by Riesz' representation theorem, such that
$$\tag{3} f(\cdot)~=~\langle u | \cdot \rangle.$$
In other words,
$$\tag{4} \langle A^{\dagger}v | w \rangle~=~\langle u | w \rangle~=~f(w)=\langle v | Aw \rangle. $$
It is straightforward to check that the adjoint operator $A^{\dagger}:H\to H$ defined this way becomes an $\mathbb{F}$-linear operator as well.
IV) Finally, let us return to OP's question and consider the definition of the adjoint of an antilinear operator. The definition will rely on the complex version of Riesz' representation theorem. Let $H$ be given a complex Hilbert space, and let $A:H\to H$ be an antilinear continuous operator. In this case, the above equations (2) and (4) should be replaced with
$$\tag{2'} f(\cdot)~=~\overline{\langle v | A(\cdot) \rangle},$$
and
$$\tag{4'} \langle A^{\dagger}v | w \rangle~=~\langle u | w \rangle~=~f(w)=\overline{\langle v | Aw \rangle}, $$
respectively. Note that $f$ is a $\mathbb{C}$-linear functional.
It is straightforward to check that the adjoint operator $A^{\dagger}:H\to H$ defined this way becomes an antilinear operator as well.
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$^{1}$We will ignore subtleties with discontinuous/unbounded operators, domains, selfadjoint extensions, etc., in this answer.
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.
Best Answer
Time reversal is not only complex conjugate, what it does is also to transpose the items on which it acts (vectors, matrices).
$$T\langle \phi|\hat{O}|\psi\rangle = \langle \psi T|\hat{O}|T \phi\rangle.$$
Notice the change of places of the functions in the right wing with respect to the left wing. Also, I used the fact that $\hat{O}$ is unchanged at time-reversal.
Now we do the following change which is allowed under the integral if the two functions vanish at infinity:
$$\langle UK\psi|\hat{O}|UK\phi\rangle = \langle \phi|U\hat{O}U^\dagger|\psi\rangle.$$
So, we got the time-reversed of $\hat{O}$.