By the double-angle formula, we have $\cos^2 x = \frac{1}{2}(1+\cos (2x))$. Using that and the addition theorem for the cosine, we find that
$$\cos^2 (\mu-\delta) - \cos^2 (\mu + \delta) = \frac{1}{2}\bigl(\cos (2(x\mu \delta)) - \cos (2(\mu+\delta))\bigr) = \sin (2\delta)\sin (2\mu).$$
Thus for points $x,y$ at a given distance $0 < h = \lvert x-y\rvert$, the difference $\lvert \cos^2 x - \cos^2 y\rvert$ is maximised if the midpoint $\mu = \frac{x+y}{2}$ satisfies $\sin (2\mu) = \pm 1$, that is, if $\mu = \frac{\pi}{4} + k\frac{\pi}{2}$ for some $k\in \mathbb{Z}$. Then we have
$$\frac{\lvert \cos^2 x - \cos^2 y\rvert}{\lvert x-y\rvert^\alpha} = \frac{\lvert\sin (x-y)\rvert}{\lvert x-y\rvert^\alpha}.$$
It is clear that for $\lvert x-y\rvert > \frac{\pi}{2}$ we have
$$\frac{\lvert \cos^2 x - \cos^2 y\rvert}{\lvert x-y\rvert} < \biggl(\frac{2}{\pi}\biggr)^\alpha = \frac{\lvert \cos^2 0 - \cos^2 \frac{\pi}{2}\rvert}{\lvert 0 - \frac{\pi}{2}\rvert^\alpha},$$
regardless of where the midpoint of $x$ and $y$ is. Hence we need to find
$$H(\alpha) := \sup \biggl\{ \frac{\sin t}{t^\alpha} : 0 < t \leqslant \frac{\pi}{2}\biggr\}.$$
Differentiating $g_\alpha(t) = t^{-\alpha}\sin t$ yields
$$g_\alpha'(t) = \frac{t\cos t - \alpha \sin t}{t^{\alpha+1}},$$
so $g_\alpha$ has a critical point where $t\cos t = \alpha \sin t$, or $t\cot t = \alpha$. Since $q(t) = t\cot t$ is strictly decreasing on $\bigl[0,\frac{\pi}{2}\bigr]$ and $q(0) = 1$ (by continuity) and $q\bigl(\frac{\pi}{2}\bigr) = 0$, for each $\alpha \in (0,1)$ there is exactly one $t_\alpha = q^{-1}(\alpha) \in \bigl(0,\frac{\pi}{2}\bigr)$ with $g_\alpha'(t_\alpha) = 0$. Since $g_\alpha'\bigl(\frac{\pi}{2}\bigr) < 0$ and $\lim\limits_{t\downarrow 0} g_\alpha'(t) = +\infty$, $g$ has a maximum at $t_\alpha$, so
$$H(\alpha) = g_\alpha(t_\alpha) = \frac{\sin t_\alpha}{{t_\alpha}^\alpha}.$$
Using $\sin t = \frac{\tan t}{\sqrt{1 + \tan^2 t}}$ for $0 \leqslant t < \frac{\pi}{2}$ and $\tan t_\alpha = \frac{t_\alpha}{\alpha}$, we obtain
$$H(\alpha) = \frac{{t_\alpha}^{1-\alpha}}{\sqrt{\alpha^2 + {t_\alpha}^2}}.$$
Unfortunately we can't express $t_\alpha$ in terms of elementary functions. It may be possible to get a closed form using the Lambert $W$-function (or some other special functions).
But it's not too difficult to numerically approximate $t_\alpha$ to any desired precision.
We note that we have $H(1) = 1$, and $H(\alpha) \leqslant 1$ for $0 < \alpha < 1$: for $0 < t \leqslant 1$ we have
$$\frac{\sin t}{t^\alpha} = t^{1-\alpha} \frac{\sin t}{t} < \biggl(\frac{\pi}{4}\biggr)^{1-\alpha} \leqslant 1,$$
for $1 < t \leqslant \frac{\pi}{2}$ we have
$$\frac{\sin t}{t^\alpha} \leqslant \frac{1}{t^\alpha} < 1.$$
In fact, we have $H(\alpha) < 1$ for $0 < \alpha < 1$, but showing the strict inequality requires a bit more work.
We can also determine the asymptotic behaviour for $\alpha \approx 0$ and $\alpha \approx 1$:
For $\alpha \approx 1$, note that
$$t\cot t = 1 - \frac{t^2}{3} - \frac{t^4}{45} + O(t^6).$$
From that we obtain
$$t_\alpha = \sqrt{3(1-\alpha)} + O\bigl((1-\alpha)^{3/2}\bigr)$$
and further
$$H(\alpha) = 1 + \frac{1-\alpha}{2} \log (1-\alpha) + O(1-\alpha).$$
For $\alpha \approx 0$, note
$$\biggl(\frac{\pi}{2} - \beta\biggr)\cot \biggl(\frac{\pi}{2} - \beta\biggr) = \biggl(\frac{\pi}{2} - \beta\biggr)\tan \beta = \frac{\pi}{2}\beta + O(\beta^2),$$
so
$$t_\alpha = \frac{\pi}{2} - \frac{2}{\pi}\alpha + O(\alpha^2)$$
and that leads to
$$H(\alpha) = \biggl(\frac{2}{\pi}\biggr)^\alpha\biggl(1 + \frac{2}{\pi^2}\alpha^2 + O(\alpha^3)\biggr) = 1 + \alpha \log \frac{2}{\pi} + O(\alpha^2).$$
In particular we see that $\lim\limits_{\alpha\to 0} H(\alpha) = \lim\limits_{\alpha\to 1} H(\alpha) = 1$.
