Assume there exists such a function $f.$
Recall that any entire function $g:\mathbb{C} \rightarrow \mathbb{C}$ satisfying the equation $g(z + \omega) = g(z)$ for all $\omega \in \Gamma$ is constant. And observe that the derivative of $f$ is of this form. It follows $f(z) = az +b \text{ for }a,b \in \mathbb{C}.$
But then
$a(z + \gamma) + b = az + b - \overline{\gamma}$ for all $\gamma \in \Gamma.$
Hence, $-a\gamma = \overline{\gamma}$ for all $\gamma \in \Gamma.$ As the $\mathbb{R}$-span of $\Gamma$ is equal to $\mathbb{C}$ and complex conjugation and complex multiplication are both $\mathbb{R}$-linear, it follows that multiplication by $-a$ and complex conjugation are equal as automorphisms of $\mathbb{C}.$ But the former is holomorphic and the latter is not. This is a contradiction. It follows no such function $f$ exists.