How does an operator transform under time reversal? We know that a time-reversal operator $T$ can be represented as
$$T=UK$$
where $U$ is some unitary operator and $K$ is the complex conjugation operator.
Then under time-reversal operation, a quantum state $|\psi\rangle$ will transform as the following:
$$|\psi^R\rangle=T|\psi\rangle=UK|\psi\rangle=U|\psi^*\rangle$$
If we require time-reversal symmetry to the system, then we need to have
$$\langle\psi^R|O^R|\phi^R\rangle=\langle\psi|O|\phi\rangle$$
where $|\psi\rangle$ and $|\phi\rangle$ are some arbitrary quantum states and $O$ is some operator. From the above equation, we would have
$$\langle\psi^*|U^{\dagger}O^RU|\phi^*\rangle=\langle\psi|O|\phi\rangle$$
So based on this equation, how do we obtain the result (given in the book "random matrices" by Mehta) that 
$$O^R=UO^TU^{\dagger}$$
where $O^T$ means the transpose of $O$.
My second question is that, what if we do NOT assume time reversal symmetry?
 A: One problem with your formula that $T$ factorizes as the product of a unitary operator $U$ and complex conjugation $K$ is that 'complex conjugation' is meaningless in a Hilbert space a priori. 
Let me be more formal about this; consider a vector $\psi \in H$, with $H$ a $n$-dimensional Hilbert space, that is, isomorphic to $\mathbb{C}^n$. How is an (abstract) vector space isomorphic to the canonical Hilbert space $\mathbb{C}^n$? In the following way: pick a basis $\{e_i\}_{i=1,...,n}$ of $H$. Then decompose $\psi$ in this basis:
$$ \psi = \sum_i \psi_i e_i \ .$$
Finally map $\psi$ on the $n$-tupel $(\psi_1,...,\psi_n)$.
As you can see, this isomorphism, lets denote it by $E$, very strongly depends on the chosen basis.
Let's ignore this anyway. On $\mathbb{C}^n$ we can define complex conjugation $K$, it is simply the map
$$ K(\psi_1,...,\psi_n) = (\overline{\psi_1},...,\overline{\psi_n}) \ . $$
Hence we can define a complex conjugation $K_E$ on $H$ simply by
$$ K_E = E^{-1} \circ K \circ E   \ .$$
Now let's see what happens if we change a basis; consider another basis $E' = \{e'_i\}_{i=1,...,n}$ with $e_i = \sum_j M_{ij} e'_j$. Then considering $M_{ij}$ as the matrix of an operation on $\mathbb{C}^n$:
$$ K_{E} = E^{-1} \circ K \circ E = E'^{-1} \circ M^{-1} \circ K \circ M \circ E' = E'^{-1} \circ  \overline{M}M^{-1} K \circ E'  \ ,$$
i.e. unless $\overline{M} = M$, $K_E \neq K_{E'}$. Hence there is no invariant notion of complex conjugation in a complex Hilbert space.
The statement is true that if we fix a basis $E$, we may write complex conjugation w.r.t. any other basis as 
$$ K_{E'} = U_{E,E'} K_E  \ ,$$
where $U_{E,E'}$ is a unitary. Note that complex conjugation depends on $E$ only through the equivalence class $[E]$ of basis connected to $E$ through real transformations.
In this language one could say that a time-reversal operation is a choice of preferred equivalence class of basis. That is, at least, if $T^2 = 1$. If $T^2 = -1$, we should map $H$ to a $\mathbb{H}^{n/2}$, where $\mathbb{H}$ are the quaternions.
This leaves you with a recipe for computing the time-reversed operator, for $T^2=1$: simply represent $O$ in a real basis (a basis invariant under $T$), then take the complex conjugate of this matrix.
Notice that neither of these manipulations depend on $T$ being a symmetry.
A: 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 $\hat{O}$.
