Does the complex conjugate of an integral equal the integral of the conjugate?

If $\int dz$ denotes a contour integral, then the answer is generally no. A correct formula is as follows:

$$ \overline{\int f(z) \; dz} = \int \overline{f(z)} \; \overline{dz}. $$

Indeed, let $\gamma : I \to \Bbb{C}$ be a nice curve parametrizing the contour $C$, then

$$ \overline{\int_C f(z) \; dz} = \overline{\int_I f(\gamma(t)) \gamma'(t) \; dt} = \int_I \overline{f(\gamma(t)) \gamma'(t)} \; dt= \int_C \overline{f(z)} \; \overline{dz}. $$

In general, answer is "no", because $$\overline{ \int f(z) dz} = \overline{\int \left( \text{Re}f(z) + i\text{Im}f(z)\right)dz}=\\ \int \overline{ \left( \text{Re}f(z) + i\text{Im}f(z)\right)(dx+i dy)}=\int\overline{{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)+ i(\text{Re}f(z)dy+\text{Im}f(z)dx)}}=\\ \int{\left(\text{Re}f(z)dx - \text{Im}f(z)dy\right)}-i \cdot\int{\left(\text{Re}f(z)dy+\text{Im}f(z)dx\right)}$$

Yet another variation: let be $\overline{\gamma}(t) = \overline{\gamma(t)}, t\in[a,b]$. Then, $$ \overline{\int_{\gamma}f(z)\,dz} = \overline{\int_a^b f(\gamma(t))\gamma'(t)\,dt} = \int_a^b\overline{f(\gamma(t))}\,\overline{\gamma'(t)}\,dt = \int_a^b\overline{f(\overline{\overline{\gamma}(t)})}\,\overline{\gamma}'(t)\,dt = \int_{\overline{\gamma}}\overline{f(\overline{z})}\,dz. $$