This ended up being partly answered in Difference between the steady state of a model and state space models? Note the comments in particular.
The Kalman filter equation is slightly different than shown above - see the comments in the other question. That is, the correction is based on the predicted X , not the previous X. To emphasize this, you could write it using your notation as
$$ \hat{X}_t = F_t \hat{X}_{t-1} +K_t ( Y_t - G_t F_t \hat{X}_{t-1}) $$
The Kalman filter gives the whole story, and is in fact the optimal estimate given certain assumptions (linearity, known covariance matrices, guassian normal distribution, or at least being representable with mean and covariance matrix). The optimality is either in terms of maximum likelihood, or as the optimal solution to a least squares (minimum variance of the estimate) solution.
The pure prediction $\hat{X}_t = F_t \hat{X}_{t-1}$ doesn't account for the recent measurements or the noise in the measurements or the comparison of the measurement noise to the process noise. The Kalman filter accounts for both the model and the measurements. It also accounts for both noise in the process model and noise in the measurements. The process noise term (V in the above notation) might represent actual physical disturbances such as wind gusts in modeling an aircraft, but typically is in effect tuned to represent some sense of the model uncertainty. As noted in the other question, the Kalman gain matrix K in effect is a form of ratio of process noise to measurement noise (the W term in the above notation). It performs a balancing act between information contained in the model and in the measurements. If you have large measurement noise, K approaches 0, and you believe the prediction, practically ignoring the measurement. If you have large process noise, K will work out so that you ignore the prediction and just use the measurements. Calculation of K involves calculating the covariance matrices of the prediction step and the correction step. See the reference in the comments in the other question.
So, the Kalman estimate is better, in fact, optimal.