No. It is true that $\mathbb{P}(X^*_T>\beta)=O(\beta^{-1})$, but you don't have a`little-o' bound. In fact it fails, and $\beta\\,\mathbb{P}(X^*_T>\beta)$ converges to a strictly positive value, precisely when X fails to be a martingale.

If S is the first time at which X hits $\beta>x$ then continuity gives 
$$
X\_{S\wedge T} = \beta 1\_{\{X^*_T>\beta\}}+1\_{\{X^*_T\le\beta\}}X_T
$$
Take expectations, and use $\mathbb{E}[X_{S\wedge T}]=x$, which follows from the fact that the first term is a local martingale stopped at time S, so is bounded (and hence a proper martingale).
$$
x=\beta\\,\mathbb{P}(X^*_T>\beta)+\mathbb{E}[1\_{\{X^*_T\le\beta\}}X_T].
$$
The final expectation converges to $\mathbb{E}[X_T]$ as $\beta$ goes to infinity, by monotone convergence. This gives
$$
\lim_{\beta\to\infty}\beta\\,\mathbb{P}(X^*_T>\beta)=x-\mathbb{E}[X_T].
$$
Now, it is a well known result that if X is a nonnegative local martingale and $X_0$ is integrable then it is a supermartingale, so $\mathbb{E}[X_T]\le\mathbb{E}[X_0]$, and equality holds precisely when it is a martingale over the range [0,T]. So, in our case, $\mathbb{P}(X^*_T>\beta)=o(\beta^{-1})$ exactly when $\mathbb{E}[X_T]=x$ and X is a martingale over the range [0,T].

An example when solutions to your SDE fails to be a martingale is $\sigma(x)=x^2$, $dX=X^2\\,dW$. The solution to this SDE can be written as $X=1/\Vert B\Vert$ for a 3-dimensional Brownian motion B started from the point $(x^{-1},0,0)$. You can calculate $\mathbb{E}[X_t]$ and determine that it is decreasing in t, so X is not a martingale - just a local martingale. This example appears in Roger's & Williams book [Diffusions, Markov Processes and Martingales][1] as an example of a local martingale which is not a proper martingale.


  [1]: http://books.google.co.uk/books?id=bDQy-zoHWfcC&lpg=PP1&dq=rogers%2520and%2520williams&pg=PP1#v=onepage&q&f=false