Faraday's induction law, that the EMF is proportional to the negative time derivative of the magnetic flux, actually describes two different effects which can be described by the same law. The first effect is based on the flux change due to the change of area of a conducting loop. In this case, the EMF is a purely magnetic force, the magnetic Lorentz force on the moving charge carriers in the conducting loop. The second effect is the induction of an electric field according to the Maxwell-Faraday equation of Maxwell's equations by a changing magnetic flux in a (not necessarily conducting) loop that is not moving relative to the magnetic field . It is, indeed, surprising that both effects can be described by the same induction law.
Let's consider what happens to charged particles in the conducting rod. I'll use a cylindrical (r, z, theta) coordinate system, with the origin located at the rod's fixed point, and the magnetic field pointing in the z-direction.
The charges in the conductor experience a $q \vec{v} \times \vec{B}$ force moving positive charges in one direction towards one end of the rod, and negative charges to the other. This charge separation sets up an electric field, and an equilibrium is reached where the net force on a given charged particle:
$\vec{F} = q \left(\vec{E} + \vec{v} \times \vec{B}\right) = 0$
Or,
$E = - \omega r B$ (where r radial position along the rod)
This results in a potential difference along the rod of:
$V = \int \vec{E} \cdot d\vec{l} = -\frac{B\omega L^2}{2}$
As this potential difference is a result of the motion of the conductor, it is referred to as a motional emf.
Now, perhaps the source of your confusion is in understanding the application of Faraday's law. You're quite right in saying the emf in any closed circuit can be related to the change of magnetic flux through that circuit according to:
$\epsilon = - \frac{d \phi}{dt}$.
I added my emphasis as in your original question there is no close circuit (loop). If we create a loop that is formed by the a line along the rod in its current position, the rod's original position, and the chord that the far end of the rod follows, then the area of that loop as a function of time is given by:
$A(t) = \frac{1}{2}L^2 \theta(t)$,
Therefore
$\epsilon = - \frac{d \phi}{dt} = -\frac{1}{2}L^2\omega B$.
Alternatively, one can calculate the motional emf as the integral around that loop of the force per unit charge (work done per unit charge):
$\epsilon = \oint \left(\vec{v} \times \vec{B}\right) \cdot d \vec{l}$,
which due to the fact the only section of the loop with non-zero velocity is the rod itself gives the same answer.
Best Answer
"I think that magnetic flux through conductor remains constant as B is constant."
It's not the flux "through the conductor" that matters. It's the flux through the area swept out by the conductor. Imagine that the straight conductor (length $\ell$) is lying on a table, and that there is a uniform magnetic field acting downwards. (Actually there is : the vertical component of the Earth's field.) You then move the conductor across the table at speed v in a direction at right angles to itself. In time $\Delta t$ it sweeps out an area $\ell v \Delta t$
The flux through the swept out area is $$\Delta \Phi = (\ell v \Delta t)B$$
So according to Faraday's law, the induced emf is $$\mathscr E=\frac {\Delta \Phi}{\Delta t}=\frac {(\ell v \Delta t)B}{\Delta t}=B\ell v$$ So we have recovered the result that you obtained from the magnetic Lorentz force. In my opinion the magnetic Lorentz force is more fundamental than Faraday's law when the emf is due to movement of conductors. However Faraday's law has the merit of spanning two types of electromagnetic induction: this one and the type due to changing flux through a stationary circuit, which depends on the electric field part of the Lorentz force.