Yes, see in Revuz-Yor exercise 4.27 chapter IV. We will apply theorem 4.10 to the the process
$$A_{T}:=\sup_{0 \leq s < t \leq T} \frac{|B_t-B_s|}{|t-s|^\alpha}.$$
It is adapted, continuous and increasing. Using this Q&A, we indeed get by Markov inequality that
$$\lim_{b\to +\infty }P_{x}(A_{\lambda^{2}}>b\lambda)=0.$$
We also have by the Markov property that
$$ \begin{split} A_{T+S}-A_{S} & \leq\sup_{S \leq s < t \leq T+S} \frac{|B_t-B_s|}{|t-s|^\alpha}\\ & =\sup_{0 \leq s < t \leq T} \frac{|B_{t+S}-B_{s+S}|}{|t-s|^\alpha}\\ & =A_{T}\circ \theta_{S}, \end{split} $$ where $\theta_{S}$ is the shift of the path (here for the cross-term $$C_{S,T+S}:=\sup_{0 \leq s\leq S \leq t \leq T+S} \frac{|B_t-B_s|}{|t-s|^\alpha}$$ we used the bound $$ \frac{|B_t-B_s|}{|t-s|^\alpha}\leq \frac{|B_t-B_{S}|}{|t-S|^\alpha}+ \frac{|B_S-B_s|}{|S-s|^\alpha}$$
and so $C_{S,T+S}-A_{S}\leq A_{S,T+S}$. Therefore) Therefore, by theorem 4.10 we get for any moderate function F (i.e. increasing, continuous, vanishing at zero and $\sup_{x>0}\frac{F(ax)}{F(x)}=\gamma<\infty$ for some $a>1$)
$$\Bbb E\left[ F\left(\sup_{0 \leq s < t \leq \tau} \frac{|B_t-B_s|}{|t-s|^\alpha}\right)\right] \leq c_{F} \Bbb E [F(\tau^{1/2})].$$
So in particular we can take $F(x)=x^{p}$ for $p\geq 1$.
