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.

Heuristically, a point $(t,B_t)$ being a extremum is antithetical to fast oscillation, since there is no oscillation on one side of (above/below) $B_t$.

However,

1. This is only a heuristic, and
2. One may wonder whether there are so many fast times and so many maxima as to force an overlap.

At least we can rule out (2) as follows:

Theorem. If $W$ and $B$ are two independent Brownian motions then the set of maxima of $W$ is disjoint from the set of fast times of $B$.

Proof: Let $M$ be the set of local maxima of $W$ 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 of $B$ is a local maximum of $W$.