If you make the statement

Fix an algebraically closed base field k and let X be a scheme of finite type over k. Then
X/k is proper iff for all smooth quasi-projective curves C/k and all maps f: C\c -> X then f extends uniquely to f': C -> X.

it seems true to me. This should basically comes down to the fact that one can classify birational equivalence classes of curves over k in terms of the 'abstract curves' coming from all possible discrete valuations on dimension 1 function fields K/k. So using the fact that in this situation it is sufficient to check the valuative criterion on DVRs it seems like it should not be so hard to see the equivalence. The argument I had in mind is as follows:

The valuative criterion implies the statement above. For the converse it is sufficient to show that any f: Spec K -> X lifts to an open subset of the curve C`_`

K determined by K/k by which I mean the unique nonsingular projective curve in the birational class corresponding to K/k (the distinguished point in the complement we think of as being removed is uniquely determined by the discrete valuation we pick so that is no problem). If f just hits a closed point we can just collapse C`_`

K via the structure map to Spec k and this is fine since there is nothing to lift. If not, we hit a dimension 1 point whose closure with the reduced induced structure determines some curve C' birational to the corresponding C`_`

K'. The map K' -> K induces a dominant morphism g: C`_`

K -> C`_`

K'. We thus get a map by taking a common (up to isomorphism) open in C`_`

K' and C' , taking its preimage U in C`_`

K and considering

U -> C' -> X
which is the desired lift of f to the quasi-projective curve U.

I think there is also a slicker argument using the categorical definition of finite presentation.

For the same reasons this works for checking separatedness when one makes the obvious modifications to the statement.

Over other bases I am not sure at the moment... I can't remember if the birational classification is still that simple (although some people implicitly mean by variety that everything is over some fixed alg. closed base).

In the more general case (if your definition of variety doesn't include a finite type over a noetherian base hypothesis) where one needs non-noetherian valuation rings I think this interpretation is false - non-noetherian valuation rings can have arbitrary dimension.