No, a Brownian motion cannot be fast at its extrema.

*Proof*: Let $M$ be the set of local maxima of Brownian motion on a given closed interval.
Then $M$ is a union of finite discrete, hence closed, sets $M_n$, where $M_n$ consists of those $t$ such that $B_s\le B_t$ for all $s$ with $|s-t|\le 1/n$.
Since $M_n$ is countable, the packing dimension $d_P(M_n)=0$.

So by Theorem 10.22 of http://research.microsoft.com/en-us/um/people/peres/brbook.pdf
we have almost surely that
$$\sup_{t\in M_n} \limsup_{h\downarrow 0} \frac{|B_{t+h}-B_t|}{\sqrt{2h\log(1/h)}}=\sqrt{d_P(M_n)}=0.$$
Given $a>0$ we call a time $t\in [0,1]$ an $a$-fast time if
$$\limsup_{h\downarrow 0} \frac{|B_{t+h}-B_t|}{\sqrt{2h\log(1/h)}}\ge a.$$
And a time is fast if it is $a$-fast for some $a>0$.
Thus, a.s. no fast time is a local maximum.