Is it possible to formulate the axiom of choice as the existence of a survival strategy?

It is certainly true that some choice is required.

An isomorphic game (mapping (giraffes, scarves, lion) to (prisoners, hats, warden)) was considered by Hardin and Taylor in their stimulating (and elementary) paper

MR2501394 Hardin, Christopher S.; Taylor, Alan D. An introduction to infinite hat problems. Math. Intelligencer 30 (2008), no. 4, 20–25.

The claim you make in your question is called by them the Gabay-O'Connor theorem (though Gabay and O'Connor do not seem to have published it).

Hardin and Taylor show that, in the case $|G| = \kappa = \aleph_0$ and $|C|=2$, it is consistent with ZF+DC (where DC is the Axiom of Dependent Choice) that the giraffes cannot win - that for every survival strategy, the lion (knowing the strategy) can assign the scarves in such a way that $\aleph_0$ many giraffes are eaten. Specifically, the existence of a winning survival strategy would imply the existence of a subset of $\mathbb{R}$ which does not have the property of Baire (BP) - and it is a celebrated theorem of Shelah that it is consistent with ZF+DC that every subset of $\mathbb{R}$ has the BP.

It can also be shown that a survival strategy would imply the existence of a subset of $\mathbb{R}$ which is not Lebesgue measurable. (One way to show this is via the Lebesgue density theorem. I also worked out a kind of cute probabilistic argument a couple years ago, which I may put in here if anyone is interested or if I get bored.) Solovay has shown (thanks Asaf for the correction) that if "ZFC + there is an inaccessible cardinal" is consistent (which I understand is widely believed to be the case), then so is "ZF + DC + every subset of $\mathbb{R}$ is Lebesgue measurable." So again we see that more than DC is needed.

On the other hand, one should be able to produce a survival strategy (in the case $|C| = 2$) from a free ultrafilter, so the ultrafilter lemma is sufficient, and I believe the ultrafilter lemma is known to be strictly weaker than AC. So the existence of a survival strategy when $|C|=2$ cannot imply AC. (Maybe a similar argument applies for other finite $C$?) In principle, it could imply some weaker yet still interesting choice principle, but I do not know whether anyone has looked at that. I don't know what I was thinking when I wrote that, but in fact I do not know of any way to get a survival strategy for this particular game from a free ultrafilter. (There are variants where a free ultrafilter suffices, e.g. if the victory condition is changed to "either all giraffes guess correctly or all guess incorrectly".)

(Disclaimer: I am not a set theorist and most of this is stuff I have picked up from folklore because it sounded interesting, so I may well have something completely wrong. The real set theorists here are more than welcome to correct me.)

Here is one interesting observation:

If the giraffes form an amorphous set, which is an infinite set all of whose subsets are either finite or co-finite, then they have a winning strategy in the case that there are only finitely many colors. The reason is that because the giraffes are amorphous, every assignment the lion might pick must be constant on a co-finite set. All the giraffes will be able to detect that one infinite color, and announce it as their own. Only finitely many will be wrong.

This shows that it is not necessary to well-order $G$ to have a winning strategy. But meanwhile, as we just explained, in any model in which $G$ is amorphous, then for any finite $C$, we do still have a selector on $C^G$, precisely because all such functions are constant except on a finite set, and so we can select the constant coloring.

The question makes sense when $\kappa=\aleph_0$ since the statement involves no choice. But according to the interesting book by Hardin and Taylor, "The mathematics of coordinated inference" (2013), mentioned above in a comment by François, even the existence of a survival strategy for $\kappa=\aleph_0$ and arbitrary $G$ and $C$ is not known to be equivalent to the Axiom of Choice. They state this as "the single most prominent open question" of chapter 3 (Question 3.6.1).

ADDENDUM

There is one reasonable reformulation of your question which avoids mentioning cardinalities (which are always tricky in a choiceless context) and which is equivalent to it in ZFC. It turns out that such a reformulation is actually a particular case of a theorem in the book that is unknown to imply AC, which would indicate that your question is also an open problem.

Fix any proper ideal $\mathcal{I}$ in the powerset of $G$. Then you can stipulate that the subset of scarves each giraffe sees wrongly is a member of $\mathcal{I}$, and ask whether there is a strategy under which only a member of $\mathcal{I}$ guesses wrong (when AC is available, the subsets of cardinality strictly less than $|G|$) would form such an ideal).

Under this reformulation, $G$ can be turned into a topological space in which the neighbourhood filter of a point consists precisely on those subsets containing the point whose complements belong to $\mathcal{I}$. For each $g \in G$ you can define an equivalence relation $\sim_g$ on $C^G$ by $f \sim_g f'$ if and only if $f$ and $f'$ agree on a deleted neighbourhood $U - \{g\}$ of $g$. Given an assignation $f$, each giraffe $g$ can detect the equivalence class of $f$ under $\sim_g$ and predict their own colour. This is precisely what Hardin and Taylor call a $\textit{near neighbourhood predictor}$. Moreover, one can easily see that all points in a given member $i \in \mathcal{I}$ are topologically isolated, and in particular any point in a nonempty subset of $i$ has a neighbourhood intersecting only finitely many points in that subset, making $i$ something which they call $\textit{weakly scattered}$. Their corollary 7.2.4 then states:

"For any space $X$ of agents and set $Y$ of colours, there exists a weakly scattered-error near neighbourhood predictor"

Once again, their list of open questions includes question 7.8.1, which asks whether the above proposition, provable in ZFC, actually implies AC over ZF.