The function described in the question is 1-Lipschitz. To argue this, we'll get an inequality in place before we start writing integrals.
If $\vert \gamma\rangle$ and $\vert\delta\rangle$ are unit vectors, then the operator $\vert\gamma\rangle\langle\gamma\vert - \vert\delta\rangle\langle\delta\vert$ has at most two nonzero eigenvalues: $\pm\sqrt{1 - \vert\langle \gamma\vert \delta\rangle\vert^2}$. We can therefore bound the $\infty$-norm of this operator in terms of the Euclidean norm for vectors like so.
$$
\bigl\| \vert\gamma\rangle\langle\gamma\vert - \vert\delta\rangle\langle\delta\vert\bigr\|_{\infty}
= \sqrt{1 - \vert\langle \gamma\vert \delta\rangle\vert^2}
= \sqrt{1 + \vert\langle \gamma\vert \delta\rangle\vert}
\sqrt{1 - \vert\langle \gamma\vert \delta\rangle\vert}\\
\leq \sqrt{2} \sqrt{1 - \operatorname{Re}(\langle \gamma\vert \delta\rangle)}
= \bigl\| \vert\gamma\rangle - \vert\delta\rangle\bigr\|
$$
(You often see a related bound for the trace norm rather than the $\infty$-norm, for which we pick up an additional factor of 2, but we're going to be interested in the $\infty$-norm instead.)
In particular, taking $\vert\gamma\rangle = U \vert\phi\rangle$ and $\vert\delta\rangle = U\vert\phi\rangle$ for any unit vector $\vert\phi\rangle$ gives this:
$$
\bigl\| U \vert\phi\rangle\langle\phi\vert U^{\dagger} - V \vert\phi\rangle\langle\phi\vert V^{\dagger} \bigr\|_{\infty} \leq
\;\bigl\| U \vert\phi\rangle - V\vert\phi\rangle\bigr\| \leq \| U - V\|_{\infty}.
$$
The $\infty$-norm is (like all norms) convex, so by thinking about the spectral decomposition of a given density operator we see that the same bound works for density operators:
$$
\bigl\| U \sigma U^{\dagger} - V \sigma V^{\dagger} \bigr\|_{\infty}
\leq \| U - V\|_{\infty}.
$$
That's the inequality we needed. Now, observing that $N(\vert\psi\rangle\langle\psi\vert)$ is a density operator for every unit vector $\vert\psi\rangle$, we get what we're after:
$$
\begin{aligned}
\bigl\vert f(U) - f(V) \bigr\vert & =
\;\Biggl\vert \int \langle \psi \vert
U N(\vert\psi\rangle\langle\psi\vert) U^{\dagger} - V N(\vert\psi\rangle\langle\psi\vert) V^{\dagger}
\vert\psi\rangle\,
\mathrm{d}\psi
\Biggr\vert\\
& \leq
\int \bigl\vert \langle \psi \vert
U N(\vert\psi\rangle\langle\psi\vert) U^{\dagger} - V N(\vert\psi\rangle\langle\psi\vert) V^{\dagger}
\vert\psi\rangle
\bigr\vert\,\mathrm{d}\psi\\
& \leq
\int \bigl\|
U N(\vert\psi\rangle\langle\psi\vert) U^{\dagger} - V N(\vert\psi\rangle\langle\psi\vert) V^{\dagger}
\bigr\|_{\infty}\,\mathrm{d}\psi\\
& \leq \int \| U - V\|_{\infty}\, \mathrm{d}\psi\\
& = \| U - V\|_{\infty}
\end{aligned}
$$
Of course you can further upper-bound this by $\|U-V\|_2$ if you wish, and we find that the function is 1-Lipschitz as claimed.
