No problem, there's \b.

  echo "one two three four five" | awk '{$3="\b"; print}'

Inserts the backspace character (^H), which you can then remove with [global] substitution:

  awk '{$3="\b"; sub(/\32\b/, ""); print}'

You can, of course, use an arbitrary sentinel value for the field to be deleted. Should work in gawk and BWK awk.

Or sticking with awk, I have this bash alias to remove excess whitespace that is just:

    awk '{$1=$1};1'