The wiki page on Mertens conjecture and the Connection to the Riemann hypothesis says
Using the Mellin inversion theorem we now can express $M$ in terms of 1/ζ as $ M(x) = \frac{1}{2 \pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} \frac{x^s}{s \zeta(s)}\, ds $ which is valid for $\color{blue}{1} < σ < 2$, and valid for $\color{red}{1/2} < σ < 2$ on the Riemann hypothesis. ... From this it follows that $ M(x) = O(x^{\color{red}{1/2}+\epsilon}) $ for all positive ε is equivalent to the Riemann hypothesis, ...
The $\color{red}{\text{red}}$ color indicates the question My question changed, due to anon's comment's, to:
If Riemann was false, would this imply a bound of $M(x)=O(x^{\color{blue}{1}+\epsilon})\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \!\! \! \! \! \! \! \! \! \! \! ---------\;\;$?$ \phantom{------------------------------------------------}$ Is $M(x)=O(x^σ)$ possible with $σ≤1$ even if RH is false?
A look at Mertens function, makes me think that it should be easy to prove this.
But I still don't have a clue...