Well, you can read a simple estimate off the proof of Kolmogorov criterion. If $(a,b)$ is a diadic interval, with $|b-a|=2^{-n}$, then, by Chebyshev's inequality, we have $$ \mathbb{P}(d(X_a,X_b)\geq M 2^{-\gamma n})\leq\frac{K2^{-n}2^{(\alpha\gamma-\beta)n}}{M^\alpha}. $$ The union bound over diadic intervals now gives $$ \mathbb{P}(\exists a,b\text{ diadic}:d(X_a,X_b)\geq M|a-b|^\gamma)\leq KM^{-\alpha}(1-2^{\alpha\gamma-\beta})^{-1}. $$ The $\gamma$-Holder norm of $X_t$ and the norm of its restriction to diadic points differ by a factor of at most $2(1-2^{-\gamma})^{-1}$. Therefore, $$ \mathbb{P}(||X_t||_\gamma\geq M)\leq K\cdot C(\alpha,\beta,\gamma)\cdot M^{-\alpha}. $$ Taking $X_t=tX_1$ with a suitable $X_1$ shows that in general, this cannot be substantially improved.

Well, you can read a simple estimate off the proof of Kolmogorov criterion. If $(a,b)$ is a diadic interval, with $|b-a|=2^{-n}$, then, by Chebyshev's inequality, we have $$ \mathbb{P}(d(X_a,X_b)\geq M 2^{-\gamma n})\leq\frac{K2^{-n}2^{(\alpha\gamma-\beta)n}}{M^\alpha}. $$ Suppose we are interested in the Hölder norm on $[0,1]$. The union bound over diadic intervals gives $$ \mathbb{P}(\exists a,b\text{ diadic}:d(X_a,X_b)\geq M|a-b|^\gamma)\leq KM^{-\alpha}(1-2^{\alpha\gamma-\beta})^{-1}. $$ The $\gamma$-Hölder norm of $X_t$ and the norm of its restriction to diadic points differ by a factor of at most $2(1-2^{-\gamma})^{-1}$. Therefore, $$ \mathbb{P}(||X_t||_\gamma\geq M)\leq K\cdot C(\alpha,\beta,\gamma)\cdot M^{-\alpha}. $$ Taking $X_t=tX_1$ with a suitable $X_1$ shows that in general, this cannot be substantially improved (EDIT: in fact, it can be slightly improved, see the answer by zhoraster).