Let $\widetilde{h}^{\ast}$ be a reduced cohomology theory that satisfies the wedge sum axiom, i.e. the cohomology of a wedge sum is the product of the cohomology groups of the summands. We pick a pointed space $(X, x_0)$ and a cohomology class $c \in \widetilde{h}^n(X)$. Then there exists a map $q$ from $\pi_1(X,x_0)$ to $\widetilde{h}^n(S^1)$ defined as follows. $$q: [f] \mapsto f^{\ast}(c)$$
What I'd like to know is whether $q$ is necessarily a group homomorphism. I think I have a proof in the case $\widetilde{h}^{\ast}$ is reduced singular cohomology, but I was wondering whether the statement is true in general.
Yes, this is true. Given $[f],[g]\in \pi_1(X,x_0)$, their composition $f*g$ is defined as the composition $S^1\stackrel{p}{\to} S^1\vee S^1\stackrel{h}{\to} X$ where $h$ is given by $f$ on one circle and $g$ on the other. We then have $$h^*(c)=(f^*(c),g^*(c))\in\widetilde{h}^n(S^1\vee S^1)\cong\widetilde{h}^n(S^1)\oplus \widetilde{h}^n(S^1)$$ since the latter isomorphism is given by the inclusions of each circle into $S^1\vee S^1$. Thus $q([f*g])=p^*(f^*(c),g^*(c))$. But $p^*(a,b)=a+b$, since $p$ is homotopic to the identity after collapsing either circle of $S^1\vee S^1$. Thus $q([f*g])=f^*(c)+g^*(c)=q([f])+q([g])$, so $q$ is a homomorphism.