Solution:

Let us assign the colors the values $0,1,2,3,4$ and denote by $B$ the color of the second prisoner's hat, and by $S$ the sum of the colors of the hats of the third through the last prisoner modulo 5. We will describe a strategy for the first two prisoners after which all the remaining prisoners will know $S$, and thus, knowing the colors of the other hats, will be able to determine their own. In this way, freedom will be guaranteed for at least 2013 prisoners. The first prisoner answers "no" if $S \in\{0, B, B+1\}$ or $(B, S)=(4,1)$. To this the second answers "no" if $S=0$, and otherwise says "yes" and gives the guard the answer $S$. In this way, if two "no" answers are heard, the others know that $S=0$. If the second says "yes" and is freed, the others know that $S=B$, while otherwise they know that $S=B+1$ if $B \neq 4$, i.e. that $S=1$ if $B=4$.

In the remaining cases the first prisoner says "yes" (it is irrelevant what happens to him). To this the second says "yes" if $S \in\{2,4\}$ and gives the guard the answer $S-2$. If $B=0$, then $S \notin\{0,1\}$, so the others know that $S=2$ if the second is freed, $S=3$ if he says "no", and $S=4$ if he says "yes" and misses. Finally, if $B \neq 0$, then in view of the affirmative answer of the first, $S \equiv B+2$ or $S \equiv B+3\ (\bmod\ 4)$ holds, and the second's answer determines the parity of $S$, which enables the others to determine $S$.

It remains to prove that no strategy guarantees the freedom of 2014 prisoners. Suppose that such a strategy exists and consider a set of five arrangements which differ only in the color of the second prisoner's hat. In at most one of these arrangements does the strategy predict that the first says "no" (otherwise the second could not determine the color of his own hat with certainty). In at least three of the remaining four arrangements the first would have to miss, so the second would not be able to determine the color of his own hat. This completes the proof.