I know that states with spin S=0 in a diatomic molecule have no spin orbit coupling, independent on the value of the projection of the total electronic angular momentum.
I expect the same is true if the absolute value of $\Lambda$ is equal to zero independent on the spin of the diatomic molecule.
Is it correct that a diatomic molecule with a $^3\Sigma$ state has no spin orbit coupling?
Yes I would think so because the projection of orbital angular momentum is zero in forming $\Omega=\Lambda + \Sigma$. This would seem to correspond to Hund's case (b) where $\Lambda =0$ but $S \ne 0$, so no spin orbit coupling (unlike Hund's case (a)(c)(d)). The spin angular momentum remains fixed in space and the molecule rotates under it. If $R$ is the whole body (molecular) angular momentum this couples with the spin so that each rotational level splits into $2S+1$ components and the total angular momentum $J$ takes values from $R+S$ to $R-S$. The splitting is scaled with a spin rotation constant rather than spin-orbit constant but in $^3\Sigma$ there is also a $s_1\cdot s_2$ spin-spin coupling term.
First, I think one needs to specify what you mean when you say that a certain electronic state, e.g. one with $S=0$, "has no spin orbit coupling".
One interpretation might be of the type: the energy of that state is unaffected by the spin-orbit operator. That is, if I start with a Hamiltonian without any spin-orbit operator and then gradually switch it on, the energy of that state is unaffected. In this strict interpretation all electronic states are affected by spin orbit, including singlet states.
In a less strict interpretation, we might merely require a weak dependence on the spin-orbit operator. In practice this usually means that, in a perturbation-theory treatment, the first-order effect of the spin-orbit operator is zero. In this sense you a right: in diatomic molecules all singlet states (any $\Lambda$) and all $\Sigma$ states (any multiplicity) have a zero first-order spin-orbit energy shift.
Finally, at a purely empirical level we might disregard spin-orbit completely if it doesn't lead to splitting of the state. If spin-orbit leads only to a (small) energy shift, it may not be experimentally measurable (not by an analysis of energy level positions, at least).
In the specific case of $^3\Sigma^\pm$ states, as I said above the effect of spin-orbit is zero at first order of perturbation, so it can be said that they "have no spin orbit coupling". At higher order, they have non-zero matrix elements with the following terms: $^{3,5}\Sigma^\mp$, $^{1,3,5}\Pi$. If a state with one of these symmetries is energetically close it'll have some effect on the $^3\Sigma^\pm$.
