This answer is along the same lines as the earlier answer by @JAlex. I suggest you consult a physics mechanics text, such as Classical Mechanics by Goldstein, for more details. I used Mathematica to do the calculations that follow.
There are two sets of coordinates to consider: the inertial (space) coordinates and the body coordinates fixed (and rotating) with the body. Using the Euler angles a vector can be expressed in either set of coordinates. The figure below provides the coordinates and the Euler angles using the convention in Goldstein.

$X_I$, $Y_I$, $Z_I$ are the inertial (space) coordinates, $X_B$, $Y_B$, and $Z_B$ are the body axes that are fixed in the body and are chosen to be the principal axes of the body. $X’$ is the line of nodes.
I will re-label the angles and axes in your figure to correspond to those in the figure above. $Z_I$ is the axle, $\theta$ is $\phi$ in your diagram, $Z_B$ is in the n direction in your diagram.
For your problem, $\theta$ is fixed , $\psi$ is zero and $\dot \phi$ is constant. In the space coordinates, $\vec \omega$ is fixed in the $Z_I$ direction: $\vec \omega = \{0,0,\omega \}$. The inertia tensor $\bf I_{body}$ in the body coordinates is as you state it. You need to transform the $\vec \omega$ from space to body coordinates.
To transform a vector $\vec V$ from space to body coordinates $\vec V_{body} = {\bf A} \vec V_{inertial}$ where $\bf A$ is the transformation matrix given in Goldstein. (To transform a vector from body to space coordinates $\vec V_{inertial} = {\bf A^T} \vec V_{body}$, where ${\bf A^T}$ is the transposed matrix.) For your case, with $\psi$ of zero ${\bf A}$ is
$\left(
\begin{array}{ccc}
\cos (\phi ) & \sin (\phi ) & 0 \\
-\cos (\theta ) \sin (\phi ) & \cos (\theta ) \cos (\phi ) & \sin (\theta ) \\
\sin (\theta ) \sin (\phi ) & \sin (\theta ) (-\cos (\phi )) & \cos (\theta ) \\
\end{array}
\right)
$
$\vec \omega_{body} = {\bf A} \vec \omega_{inertial}$ = $\{0,\omega \sin (\theta ),\omega \cos (\theta )\}$
$\vec L_{body} = {\bf I_{body}} \vec \omega_{body}$ = $\left\{0,\frac{1}{4} m R^2 \omega \sin (\theta ),\frac{1}{2} m R^2 \omega \cos (\theta )\right\}$
$\vec L_{space} = {\bf A^T} \vec L_{body}$ =$\left\{\frac{1}{4} m R^2 \omega \sin (\theta ) \cos (\theta ) \sin (\phi ),-\frac{1}{4} m R^2 \omega \sin (\theta ) \cos (\theta ) \cos (\phi ),\frac{1}{4} m R^2 \omega \sin ^2(\theta )+\frac{1}{2} m R^2 \omega \cos ^2(\theta )\right\}$
Since the object is rotating about an axis $Z_I$ that is not a principal axis, torque is required to maintain $\vec \omega$ fixed along $Z_I$.
The torque $\vec N$ is given by the Euler equations (see Goldstein).
For constant $\vec \omega$, $\vec\omega \times ({\bf I} \cdot \vec \omega) = \vec N$.
$\vec N_{body} = \left\{\frac{1}{4} m R^2 \omega ^2 \sin (\theta ) \cos (\theta ),0,0\right\}$.
$\vec N_{space} = \left\{\frac{1}{4} m R^2 \omega ^2 \sin (\theta ) \cos (\theta ) \cos (\phi ),\frac{1}{4} m R^2 \omega ^2 \sin (\theta ) \cos (\theta ) \sin (\phi ),0\right\}$
You can also transform the inertia tensor between body and inertial coordinates.
$\bf I_{inertial} = \bf A^T \cdot I_{body} \cdot A =$
$\tiny \left( \begin{array}{ccc}
\frac{1}{2} m R^2 \sin ^2(\theta ) \sin ^2(\phi )+\frac{1}{4} m R^2 \cos ^2(\theta ) \sin ^2(\phi )+\frac{1}{4} m R^2 \cos ^2(\phi ) & -\frac{1}{4} m R^2 \cos ^2(\theta ) \sin (\phi ) \cos (\phi )-\frac{1}{2} m R^2 \sin ^2(\theta ) \sin (\phi ) \cos (\phi )+\frac{1}{4} m R^2 \sin (\phi ) \cos (\phi ) & \frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \sin (\phi ) \\
-\frac{1}{4} m R^2 \cos ^2(\theta ) \sin (\phi ) \cos (\phi )-\frac{1}{2} m R^2 \sin ^2(\theta ) \sin (\phi ) \cos (\phi )+\frac{1}{4} m R^2 \sin (\phi ) \cos (\phi ) & \frac{1}{4} m R^2 \cos ^2(\theta ) \cos ^2(\phi )+\frac{1}{2} m R^2 \sin ^2(\theta ) \cos ^2(\phi )+\frac{1}{4} m R^2 \sin ^2(\phi ) & -\frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \cos (\phi ) \\
\frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \sin (\phi ) & -\frac{1}{4} m R^2 \sin (\theta ) \cos (\theta ) \cos (\phi ) & \frac{1}{4} m R^2 \sin ^2(\theta )+\frac{1}{2} m R^2 \cos ^2(\theta ) \\
\end{array}
\right)$
$\bf I_{body} = \bf A \cdot I_{inertial} \cdot A^T = \left(
\begin{array}{ccc}
\frac{m R^2}{4} & 0 & 0 \\
0 & \frac{m R^2}{4} & 0 \\
0 & 0 & \frac{m R^2}{2} \\
\end{array}
\right)$, your original inertia tensor in body coordinates using the principal axes.