When you increase the temperature such that change in Gibb's free energy becomes zero, does Keq = 1 at that temperature?
Yes and no.

$K = 1$ is true, and it follows directly from the equation $\Delta G^\circ = -RT \ln K$.

However, there is no reason for it to be "an equilibrium in which reactants and products are in standard conditions". $K = 1$ does not necessarily mean that the concentrations (or activities if you prefer) of all reactants and products are also equal to $1~\mathrm{mol~dm^{-3}}$ (or $1$, in the case of activities). Remember that $K$ is a fraction. For the reaction:

$$\nu_{r1}\ce{R1} + \nu_{r2}\ce{R2}\cdots \longrightarrow \nu_{p1}\ce{P1} + \nu_{p2}\ce{P2} \cdots$$

we have:

$$K = \frac{[\ce{P1}]^{\nu_{p1}}[\ce{P2}]^{\nu_{p2}}\cdots}{[\ce{R_1}]^{\nu_{r1}}[\ce{R_2}]^{\nu_{r2}}\cdots}$$

Just like how the fractions $\displaystyle \frac{2}{2}$, $\displaystyle \frac{4}{4}$, or $\displaystyle \frac{4000\hbar\pi e}{4000\hbar\pi e}$ are all equal to $1$, the individual concentrations $[\ce{P1}], [\ce{P2}], [\ce{R1}], [\ce{R2}]\cdots$ might not be equal to $1~\mathrm{mol~dm^{-3}}$, but the result will still be $1$.