- Does it need to be at $25~^\circ\mathrm{C}$?
No. $\Delta G^\circ$ is by definition independent of temperature.
- Does $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ always use $T = 298~\mathrm{K}$?
No. You use whatever temperature you are running your reaction at.
- (...)
You got it correct up to a certain point. Yes, at equilibrium, $\Delta G = 0$ and $Q = K$. However, everything after that is wrong. There is no constraint that $\Delta G = \Delta G^\circ$. They are related by the equation
$$\Delta G = \Delta G^\circ + RT\ln Q$$
$Q$ is not necessarily equal to $1$, so $\ln Q$ is not necessarily $0$ and $\Delta G$ is not necessarily equal to $\Delta G^\circ$.