$\hat{p}$ is Hermitian and Hermitian operators $O$ satisfy, by definition,
$$\hat{O} = \hat{O}^\dagger$$
Adjoint is not a synonym for complex conjugate. $\hat{p} = -i\hbar \nabla \rightarrow +i\hbar \nabla^\dagger \rightarrow -i\hbar \nabla =\hat{p}^\dagger$, but $\hat{p} \neq \hat{p}^*$.
If you wish to use Dirac notation then you should not put operators inside either bra or ket symbols. You can write
$$
\hat{Q} | \phi \rangle
$$
and
$$
\langle \psi | \hat{Q} | \phi \rangle
$$
and
$$
( \hat{Q} | \phi \rangle )^\dagger = \langle \phi | \hat{Q}^\dagger
$$
but the notation
$$
\langle \hat{Q}^\dagger \phi | \;\;\;\;\;\;\;\;\;????
$$
is an abuse of Dirac notation. Of course you can write it if you want to adopt some other notation: that is up to you. You could, for example, announce that this quantity shall be defined, in your own work, to be identical to
$$
\langle \phi | \hat{Q}
$$
which appears to be what you are doing in your opening formula. But in Dirac notation you should not confuse matters by putting operators inside bra symbols or ket symbols. Inside the bra or ket should be simply a label: a symbol to say which bra or ket it is. The operator then acts on the whole thing. That is, $\hat{Q}$ acts on the ket $| \psi \rangle$ not on the label $\psi$.
You may be unfamiliar with writing an operator on the right of a bra, or as the last part of an expression. But it is in fact entirely logical and maps correctly to the mathematical behaviour of these mathematical objects. If in doubt, consider the case where an operator can be represented by a matrix and a ket can be represented by a vector.
[Final remark to pre-empt comments disagreeing with the above.
Mathematical notation does not have to be Dirac notation, so if you have seen something like $| \hat{Q} \psi \rangle$ or $\langle \hat{Q} \psi|$ in a reputable text, it does not mean I am saying that text is wrong, I am just saying that at this point it departs from Dirac notation, and I would not recommend someone learning the subject to depart from well-formed Dirac notation.]
Further note
A much-used fact about kets and their decomposition is
$$
| \psi \rangle = \sum_n | n \rangle \langle n | \psi \rangle
$$
where $\{ |n \rangle\}$ is a complete orthonormal set of kets (also called state-vectors). By repeatedly using this one finds things like
$$
\langle \phi | \hat{Q} | \psi \rangle
= \sum_n \sum_m
\langle \phi | n \rangle \langle n | \hat{Q}
| m \rangle \langle m | \psi \rangle .
$$
It follows that if $v_\phi$ is the vector whose components are $\langle n | \phi \rangle$ and $v_\psi$ is the vector whose components are $\langle n | \psi \rangle$ and
$Q$ is the matrix whose components are $\langle n | \hat{Q} | m \rangle$ then
$$
\langle \phi | \hat{Q} | \psi \rangle
= v_\phi^\dagger Q v_\psi
$$
where the expression on the right is written in the standard way for products of vectors and matrices.
For any such vectors and matrix it is the case that
$$
v_\phi^\dagger Q v_\psi
= (Q^\dagger v_\phi)^\dagger v_\psi
$$
and I expect this explains why someone was proposing to move the $Q$ operator in front of the $\phi$ and put a dagger on it in the incorrectly-formed expression which the question asked about.
A similar fact arises for continuous variables.
In the case of a continuous set of states the sums become integrals, and if $| x \rangle$ is a position eigenstate then
the quantity $\langle x | \psi \rangle$ is commonly written $\psi(x)$ and called a wavefunction. Here the letter $\psi$ is serving first as a label on a ket, and then as the name of a function. In this case the above expression becomes, for example,
$$
\langle \phi | \hat{Q} | \psi \rangle
= \int_{-\infty}^\infty \phi^*(x) Q \psi(x) dx
$$
where $Q$ is now an operator acting on functions of $x$. And we can, if we wish, now use
$$
\langle \phi | \hat{Q} | \psi \rangle
= \int_{-\infty}^\infty (Q^\dagger \phi(x))^\dagger \psi(x) dx.
$$
One can thus regard an operator either as acting to the right or as its adjoint acting to the left.
Best Answer
Indeed, you missed that $\sigma_{00}^0=I$.
\begin{equation} \hat{U}=I+\sigma_{00}(e^{i\omega}-1), \hspace{6 mm} \hat{U}^{\dagger}=I+\sigma_{00}(e^{-i\omega}-1), \end{equation}
and it is now easy to see that $\hat{U}^{\dagger}\hat{U}=I$.