Bounds for the value of zeta take the form $$ |\zeta(\sigma +it) | \ll t^{ f(\sigma)+\epsilon} $$ where $|t|>2$ and $f$ is some function of $\sigma$. We can take $f(1)=0$ and $f(0)=1/2$ and we can always take $f$ to be convex.
Any bound of this form for $\zeta$ implies an identical form for the derivative of $\zeta$: We can express the derivative of $\zeta$ as the integral of $\zeta$ over a circle and taking the radius of this circle to go to zero slowly the loss between the bound for $\zeta$ and the bound for $\zeta'$ goes to $0$.
There can't be a better bound for $\zeta'$ at an arbitrary point than $\zeta$, since we can also express $\zeta$ as an integral of $\zeta'$. There could conceivably be a better bound for $\zeta'(\rho)$ specifically at zeroes $\rho$ of $\zeta$ but this seems unlikely - we don't typically expect the derivative of a holomorphic function to be especially small at its zeroes.
In particular, if you have no information on the real part of the zero, then the worst case is when $\sigma$ is close to $0$ so $f$ is close to $1/2$ and the best bound will be of the form $t^{1/2+\epsilon}$. Zero-free regions are not currently sufficient to rule out this probably.