Max-cut equivalence with most likely assignment to an Ising model
Finding the most likely assignment in the Ising model is equivalent to maximum cut, so forget about minimum cut for a minute.

In the formulation you give for the Ising model, we are trying to maximize $\sum_{ij\in E}\lambda_{ij}x_{ij}$, where $x_{ij}=1$ if $x_i=x_j$ and $x_{ij}=-1$ otherwise.  In maximum cut, we take each edge weight $w$ and try to maximize $\sum_{ij\in E}w_{ij}(-x_{ij})$ -- this is equivalent to maximizing the weight of the cut because the total weight remaining is completely determined by the cut.  So we just want to set $w_{ij}=\lambda_{ij}$ to make an equivalent formulation.

For minimum cut, the problem doesn't make any sense unless you have fields in the Ising model, either local or global.