Let $(\Omega, \mathcal{F}, \mathbb{P}, \mathcal{F}_t)$ be a given
probability space with usual conditions, on which $W$ is a standard
Brownian motion. For $x \ge 0$, consider
  $$X(t) = x + \int_0^t \sigma (X(s)) dW(s)$$
Assume $\sigma \in C^{0,1/2}_{loc}$, $\sigma(0) = 0$, $\sigma>0$ on  $(0,\infty)$.
By [Karatzas and Shereve 98], there exists a unique strong solution with
absorbing state at zero. Denote the running maximum by $X^*(T) =
\sup_{s\in [0,T]} X(s)$.

Question: For a fixed $T$, is this possible to show that
$\mathbb{P} ( X^*(T) \ge \beta) = o(\beta^{-1})$ as $\beta \to \infty$?

I am trying to use time-changed Brownian motion, i.e. $X(t) = x +
B([X]_t)$, where $B$ is BM, and $[X]$ is quadratic variation. There is
also density function available for running maximum $B^* (T)$, i.e. 
$\mathbb{P}(B^*(T) \ge \beta) = 2 - 2 \Phi(\beta/\sqrt{T}) =
o(\beta^{-1})$, where $\Phi(\cdot)$ is c.d.f of standard normal
distribution. But, I could not succeed using those facts to prove it. 

Thank you for your time.