Yes, this is true, in the sense that if one side is finite then so is the other and they are equal. All you really need is to notice that $\|g\|_q = \sup \{{\rm Re}\int fg: \|f\|_p = 1\}$ (since you can multiply any $f$ by a scalar of modulus 1).
I'll show that $\|A^*\|_{L^q \to L^q} \leq \|A\|_{L^p\to L^p}$; the reverse inequality follows by symmetry. Choose $f$ and $g$ with $\|f\|_p = \|\bar{f}\|_p= \|g\|_q = \|\bar{g}\|_q =1$ and $$\langle Af, g\rangle = \langle f, A^*g\rangle = {\rm Re}(\int f\cdot \overline{A^*g}) \geq \|A^*\|_{L^q\to L^q} - \epsilon.$$$$\langle Af, g\rangle = \langle f, A^*g\rangle = {\rm Re}\left(\int f\cdot \overline{A^*g}\right) \geq \|A^*\|_{L^q\to L^q} - \epsilon.$$ (By truncating, we can assume $f$ and $g$ both lie in $L^2(U)$.) That is, $\int Af\cdot\bar{g} \geq \|A^*\|_{L^q\to L^q} - \epsilon$, which shows that $\|A\|_{L^p\to L^p} \geq \|A^*\|_{L^q\to L^q} - \epsilon$.
