Here's a counterexample: take $G=S^1=\{z\in\mathbb{C}:|z|=1\}$,
$$
f(z)=\begin{cases}z^2:\mathrm{Im}(z)\geq 0,\\
\overline{z}^2:\mathrm{Im}(z)\leq 0.
\end{cases}
$$
This $f$ is nullhomotopic, but is an odd map because $\int_G \varphi\circ f=\int_G \varphi$ for all $\varphi:G\to\mathbb{C}$.
In the example above, $f$ is continuous, but not differentiable at $z=\pm 1$. The result is true for $C^1$ functions.
Suppose $f:G\to G$ is a continuously differentiable odd map, and for simplicity assume $G$ is connected. We have $\int_G \varphi\circ f\,d\mu=\int_G \varphi\,df_*(\mu)$, where $f_*\mu$ is the pushforward measure. Since $\int_G \varphi\,d\mu=0\Rightarrow\int_G\varphi\,df_*\mu=0$, we must have that $f_*\mu$ is a scalar multiple of $\mu$, and therefore $f_*\mu=\mu$ because $\mu$ has finite total measure.
For $g\in G$, let's define the Jacobian determinant of $f$ at $g$ to be the determinant of the map induced on $T_g(G)$ by $L_{gf(g)^{-1}}\circ f$, and let's write $J_f(g)$ for this determinant. If $g$ is a regular value of $f$, then there is a neighborhood $U$ of $g$ which is evenly covered by $f$, say $f^{-1}(U)=V_1\sqcup\ldots\sqcup V_k$, and on $U$
$$
f_*\mu = \sum_{i=1}^k\frac{1}{|J_f\circ (f|_{V_i})^{-1}|}\mu,
$$
so that $f_*\mu=\mu$ implies that on $U$:
$$
\sum_{i=1}^k\frac{1}{|J_f\circ (f|_{V_i})^{-1}|}=1
$$
on $U$.
This should imply that $J_f$ doesn't vanish anywhere, and since we assume $G$ is connected, $J_f$ must have constant sign. For any $g\in G$, we can compute the action of $f$ on the top homology group of $G$ locally, and is is multiplication by $\pm(\#f^{-1}(g))$ (where the $\pm$ is the sign of $J_f$). In particular this value is non-zero for some (hence all) $g\in G$. Since $f$ induces a non-zero map on the top homology of $G$, $f$ is not null-homotopic.
