Gibbs free energy in standard state vs. equilibrium
As explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$\Delta G=\Delta G^\circ+ RT\ln{1}=\Delta G^\circ$$ On the other hand at equilibrium $Q=K$ and so $$\Delta G=\Delta G^\circ + RT\ln{K}$$ This of course leads to $\Delta G^\circ = -RT\ln{K}$ since at equilibrium $\Delta G=0$. 

So you might want to think of it as multiple statements:


For the conversion of reactants to products in their standard states $Q=1$ 
At equilibrium  $\Delta G=0$
At equilibrium $Q=K$
$\Delta G^\circ$ is the free energy change for conversion of reactants to products in their standard states. 


The first statement is consistent with the definition of standard states.
The second and fourth statements follow from combination of the first and second laws of thermodynamics.
The third statement is a definition of $K$.

$K$ and $\Delta G^\circ$ are very much connected, but $\Delta G^\circ$ does not describe the change in free energy from reactants to products at equilibrium. $K$ and $\Delta G^\circ$ are not points on the reaction progress curve. Rather they are related to the slope of that curve.