I don't follow the question completely; in particular, the expression $\newcommand{\KL}{\mathrm{KL}}$ $\KL(\hat \mu \| \mu)$ does not make sense to me because $\mu$ is a real number rather than a distribution. (Edit: The question is now edited.)
Nevertheless I will mention that Sanov's theorem seems to be related to what the OP has in mind. Suppose that the true distribution is $q$ and the empirical distribution is $\hat p$. Sanov's theorem allows us to measure the KL divergence between $\hat p$ and $q$; in fact, it does a bit more, as we will see. (I am quoting the theorem from the wikipedia page.)
Sanov's theorem. Let $A$ be a set of probability distributions over an alphabet $X$, and let $q$ be an arbitrary distribution over $X$ (where $q$ may or may not be in $A$). Suppose $x_1, x_2, \ldots, x_n$ are $n$ i.i.d. samples from $q$. Then the probability that the empirical distribution of the sample lies inside $A$ is at most $(n+1)^{|X|} 2^{-n \delta}$, where $ \delta := \inf_{p \in A} \ \KL(p \| q) .$
The OP's question then follows as a
Corollary. If $\hat p$ is the empirical distribution, then $ \Pr[\KL(\hat p \| q) \geqslant \delta] \leqslant (n+1)^{|X|} 2^{- n \delta}. $
Proof. Apply Sanov's theorem with $A$ defined as the set of distributions $p$ such that $\KL(p \| q) \geqslant \delta$.