There is a specific sort of situation I know about where that sign matters. Suppose you have $f:X \rightarrow Y$ and $g:Z \rightarrow W$ cofibrations (if the maps are not cofibrations, all the same things work - you just replace the quotient spaces by mapping cones). You extend both maps to their Dold-Puppe sequences, so you get the sequences
$X \rightarrow Y \rightarrow Y/X \rightarrow \Sigma X \rightarrow \Sigma Y \ldots$
and
$Z \rightarrow W \rightarrow W/Z \rightarrow \Sigma Z \rightarrow \Sigma W \ldots$
Now suppose you have maps $a: X \rightarrow W$ and $b: Y \rightarrow W/Z$ making the obvious square commute up to homotopy. You can then extend these to make a commutative ladder from the first Dold-Puppe sequence to the second. (Notice that the sequences are deliberately offset from each other by one spot.)
Using the usual parameters and the obvious choices of homotopies you will get a square involving $Y/X, \Sigma X, \Sigma Z, \Sigma W$. This square will commute if it includes the map $-\Sigma g: \Sigma Z \rightarrow \Sigma W$, but not generally with the map $\Sigma g$. (To check all this, I recommend doing the Dold-Puppe sequences with mapping cones rather than quotient spaces but keeping the homotopy equivalences with the quotient spaces in mind, which is the only way I know to calculate what the right maps should be.)
At this point, if you were feeling stubborn, you could replace the map in your ladder $\Sigma X \rightarrow \Sigma Z$ with $-1$ times that map, and that would allow you to have used $\Sigma g$ in the square I mention in the above paragraph, but that creates other issues; if you choose not to simply use suspensions of your original maps to go from one Dold-Puppe sequence to the other then you run into problem when you are mapping between Dold-Puppe sequences without the shift of this example.
I hope this helps unravel Greg's answer (which is correct - you need the sign to get good mapping properties).
Of course one sees exactly the same phenomenon in the category of chain complexes of abelian groups (where homotopy is chain homotopy) and other such categories. I agree with Theo and Mark that one thinks about the suspension as "odd" (in the sense of parity not the sense of peculiar).
The published paper that Mark refers to that has an error of exactly this sort (which is unfortunately fundamental to the paper) is by Lin Jinkun in Topology v. 29, no. 4, pp. 389-407. I read this paper in preprint form in 1988 and missed this error, but discovered it in 1992 when reading another paper by the same author with the same error. In the Topology paper the error is made in diagram 4.4 on the right hand square (proof of Lemma 4.3).
Here is a candidate space for Cech homology. (which doesn't satisfy the disjoint union axiom, and so strictly speaking it's not a homology theory)
For any natural number n, let $X_n$ be the circular arc of radius 1 + 1/n centered at $( 1+1/n,0)$ and extending between angles $-\pi \leq \theta \leq (\pi - 1/n)$. Let $X$ be the union of these circles with the subspace topology, which is path connected.
Covering this space with sufficiently small disks (each $X_n$ needing to be covered by progressively smaller disks) gives a Cech nerve homotopy equivalent to a similar space where all but finitely many of the circular arcs have been closed up to circles. The zero'th homology of this cover is $\mathbb{Z}$ and the first homology is an infinite direct sum $\oplus_{n \geq N} \mathbb{Z}$. All other homology groups are zero.
As you decrease the size of the cover, you get a cofinal sequence of open covers inducing a decreasing sequence of abelian groups as N grows. There is a resulting exact sequence
$$
0 \to lim^1(\oplus_{n \geq N} \mathbb{Z}) \to \check{H}_0(X) \to lim^0(\mathbb{Z}) \to 0
$$
and the left-hand side is $(\prod \mathbb{Z}) / (\oplus \mathbb{Z}) \neq 0$.
Best Answer
For homology theories on CW-complexes or homology theories that map weak equivalences to isomorphisms, that's Brown's representability theorem, which you can find in any textbook on stable homotopy theory. You forgot the important axiom of excision, by the way. The short answer is yes.