My solution would be as follows:

ClearAll[ContainsAllMultiplicities]
ContainsAllMultiplicities[l1_List, l2_List] := 
 Module[{c1, c2, m, il1, il2},
  c1 = {1, #} & /@ Counts[l1];
  c2 = {2, #} & /@ Counts[l2];
  m = Merge[{c1, c2}, Identity];
  m = Values[m];
  If[MemberQ[m, {{2, _}}],
   False
   ,
   m = DeleteCases[m, {{1, _}}];
   m = SortBy[First] /@ m;
   m = m[[All, All, 2]];
   AllTrue[m, GreaterEqual @@ # &]
   ]
  ]

Some test code:

ContainsAllMultiplicities[{a, b, c}, {c, b}]
ContainsAllMultiplicities[{a, b, c}, {c, c}]
ContainsAllMultiplicities[{a, b, c}, {c, c, d}]
ContainsAllMultiplicities[{a, b, c, c}, {c, a, c}]
ContainsAllMultiplicities[{a, b, c, c}, {c, a, c, b}]
ContainsAllMultiplicities[{a, b, c, c}, {c, a, c, b, c}]

returning:

True
False
False
True
True    
False

Should work fast, especially for large lists…

But you erroneously give containsQ[{a,b,c,b},{b}] -> False

Sorry about that. I misinterpreted the requirement

observing multiplicities

to mean that the multiplicities in the superset/subset lists must match.

ContainsAll does not consider multiplicities so maybe

containsQ[l1_List, l2_List] := ContainsAll[Tally[l1], Tally[l2]];

Note that ContainsAll[list, {}] always returns True.