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 as an example of a local martingale which is not a proper martingale.

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 as an example of a local martingale which is not a proper martingale.