Does Herbrand's theorem mean any first-order logic formula can be expressed in CNF?
Well, that's not really what Herbrand's theorem says.  That's not a good way to think about Herbrand's theorem.  To understand the answer to your question, you first need to understand a little better what Herbrand's theorem does and doesn't imply.

I suspect I know what you are thinking.  You are thinking, given a statement $\varphi$ of first-order logic, you can find a disjunction $\psi_1 \lor \dots \lor \psi_k$ of quantifier-free formulas that is equivalent to $\varphi$.  However, that's not really what Herbrand's theorem says.  Herbrand's theorem doesn't necessarily give an effective or easy way to find the disjunction.  Herbrand's theorem says that if $\varphi$ is true, then there exists such a disjunction, but to actually find the disjunction, you have to know that $\varphi$ is true, and you have to know a proof that it is true (roughly speaking).

Let me give you an example and an analogy that may clear things up.  Here is D.W.'s theorem:

D.W.'s Theorem. Any formula of first-order logic can be expressed as either True or False.  In other words, for every sentence $\varphi$ of first-order logic, there is a formula $\psi \in \{\text{True}, \text{False}\}$ such that $\varphi$ is valid if and only if $\psi$ is valid.

D.W.'s theorem is trivially true; I can prove it in a single line.  Pretty cool, huh?

However, D.W.'s theorem is actually not very useful.  Given a formula $\varphi$ of first-order logic, D.W.'s theorem doesn't help you figure out whether $\varphi$ is valid or not.  Similarly, Herbrand's theorem doesn't really help you figure out whether $\varphi$ is valid or not.

Now, the formula $\text{True}$ is in Conjunctive Normal Form.  Also, the formula $\text{False}$ is in Conjunctive Normal Form.  Therefore, by D.W.'s theorem, every sentence $\varphi$ of first-order logic is equivalent to some quantifier-free formula $\psi$ in CNF form.  However, this probably not the kind of equivalence you were hoping for, because given $\varphi$, it's not necessarily easy to compute the $\psi$ that is equivalent to $\varphi$.