Yes. Actually, $x^*Ax\in\mathbb R$ for all $x\in\mathbb C^n$ is enough to force $A$ to be hermitian.
Given $\alpha\in\mathbb C$, we have $ (x+\alpha y)^*A(x+\alpha y)=x^*Ax+|\alpha|^2y^*Ay+\overline\alpha y^*Ax+\alpha x^*Ay. $ As the term on the left is real, we have that the right hand side equals its conjugate: $ x^*Ax+|\alpha|^2y^*Ay+\overline\alpha y^*Ax+\alpha x^*Ay=x^*A^*x+|\alpha|^2y^*A^*y+\alpha x^*A^*y+\overline\alpha y^*A^*x. $ As $x^*Ax, y^*Ay\in\mathbb R$, we have $x^*A^*x=x^*Ax$, $y^*A^*y=y^*Ay$. Then $ \overline\alpha y^*Ax+\alpha x^*Ay=\alpha x^*A^*y+\overline\alpha y^*A^*x. $ Taking $\alpha=1$, we get $\tag{1} y^*Ax+x^*Ay=x^*A^*y+y^*A^*x. $ Taking $\alpha=i$, we get $\tag{2} -y^*Ax+x^*Ay=x^*A^*y-y^*A^*x. $ Subtracting $(2)$ from $(1)$, we get $ y^*Ax=y^*A^*x. $ As $x,y$ we arbitrary, we conclude that $A=A^*$.
Regarding your second question, there is no agreement. Some require positive-definite matrices to be hermitian, while others simply ask for $x^*Ax>0$. The two things are the same in the complex case, but not in the real one.