In the case of the reals, the uncountable version reduces easily to the countable version, since there will be a countable subfamily of the given uncountable family that is cofinal in the inclusion order of the intervals. Basically, one can find a countable subfamily with the same intersection.
But the nested interval property (allowing arbitrary families) is interesting in a general linear order, where one cannot always find such countable subfamilies. In this context, the nested interval property is not equivalent to the completeness of the order. This can be seen by considering the order $\omega+\omega_1^*$, that is, the order consisting of an increasing $\omega$-sequence with a decreasing $\omega_1$ sequence on top of it. This linear order is not complete, since the initial $\omega$-sequence has no least upper bound, but it has the nested interval property for arbitrary nested sequences of intervals as a result of the mis-match in the cofinality of the gap. That is, any nested sequence of intervals $[a_\alpha,b_\alpha]$ will have nonempty intersection if the interval is eventually on one side of the gap, by compactness, and if the intervals straddle the gap, then if the sequence has countable cofinality, then the $b_\alpha$'s will be bounded in the $\omega_1^*$ part of the order, leading to a nonempty intersection; and if the sequence has uncountable cofinality, then the $a_\alpha$'s must eventually stabilize, and so again the intersection will be nonempty.
Meanwhile, the countable-NIT for ordered fields does not imply completeness, since if $F$ is, say, a nonprincipal ultrapower of the real field, then $F$ will be saturated for countable types, and the type of "being inside the intervals of a given countable nested sequence" is finitely consistent, hence realized inside the model. So any such $F$ satisfies the countable-NIT.
I am unsure about the general situation of NIT in incomplete ordered fields.