Generate a random pattern with at-most n repeated consecutive elements

How about this?

rand[sample_, size_, n_] := Module[{
   rs = RandomChoice[sample, RandomInteger[{0, n}]],
   rand0},
  rand0 = 
   NestList[RandomChoice[Delete[sample, #]] &, RandomChoice[sample], 
    size - 1 - Length[rs]];
  Fold[Insert[#1, #2, RandomChoice[Position[#1, #2]]] &, rand0, rs]
  ]

Update: Borrowing part of swish's answer to generate consecutive-free samples and using it with Part assignment (rand2) and ReplacePart (rand3):

ClearAll[consecutiveFreeRandom , rand2,rand3]
consecutiveFreeRandom[sample_,size_]:= NestList[RandomChoice[Delete[sample, #]] &,
    RandomChoice[sample], size-1]  

rand2[sample_, size_, n_] := Module[{rpt = RandomInteger[{0, n}], pos, rand0},
  pos = RandomSample[Range[size - rpt], rpt] ;
  rand0 = consecutiveFreeRandom[sample, size  - rpt];
  rand0[[pos]] = Transpose[{rand0[[pos]], rand0[[pos]]}]; Flatten @ rand0 ] 

rand3[sample_, size_, n_] := Module[{rpt = RandomInteger[{0, n}], pos, rand0},
  pos = RandomSample[Range[size - rpt], rpt] ;
  rand0 = consecutiveFreeRandom[sample, size - rpt] ;
  rand0 = ReplacePart[rand0, # -> Sequence[rand0[[#]], rand0[[#]] ]& /@ pos]] 

Examples:

SeedRandom[1]
rand2[Range@3, 20, 2]

{1, 2, 2, 3, 1, 3, 2, 3, 1, 3, 2, 1, 2, 1, 3, 2, 2, 1, 3, 2}

SeedRandom[1]
list = CharacterRange["A", "C"];
list[[rand2[Range @ Length @ list, 20, 2]]]

{"C", "A", "A", "B", "A", "B", "A", "B", "C", "B", "C", "A", "B", "A", "B", "A", "B", "A", "B", "C"}

SeedRandom[1]
rand3[Range@2, 30, 5] 

{1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2}

By selecting positions up to the allowed number of consecutives from the positions in the consecutive-free sample and repeating the elements in those positions, (1) rand2 and rand3 avoid the potential error that rand produces (when Position[#, #2] in its last line is {}) and (2) they are faster than rand when the bound on the number of consecutives is large.

timings = ArrayReshape[Table[{i, ss, rp, 
    First[RepeatedTiming[Quiet@rand[Range[i], ss, rp]]] , 
    First[RepeatedTiming[rand2[Range[i], ss, rp]]] , 
    First[RepeatedTiming[rand3[Range[i], ss, rp]]] }, 
  {i, {5, 100}}, {ss, {1000, 10000}}, {rp, {0, 1, 10, 500}}], {16, 6}];

timings[[All,{4, 5, 6}]] = NumberForm[#, {7, 5}]& /@ # & /@ timings[[All,{4, 5, 6}]]; 

Join[{{"input", "sample", "max number of", "timings", SpanFromLeft, 
    SpanFromLeft}, {"range", "size", "consecutives", 
    Item["rand", Frame -> True], Item["rand2", Frame -> True], 
    Item["rand3", Frame -> True]}}, table] // 
 Grid[#, Alignment -> Center, Dividers -> {All, {Range[25], {2 -> None}}}] & 

Original answer:

ClearAll[f]
f = Module[{}, While[Count[Differences[rc = RandomChoice[#, #2]], 0] > #3,
     rc = RandomChoice[#, #2]]; rc] &;

Alternatively, you can use Total[1 - Unitize[Differences[rc = RandomChoice[#, #2]]]] > #3 in place of Count[Differences[rc = RandomChoice[#, #2]], 0] > #3.

SeedRandom[1]
f[Range@4, 10, 0]

{2, 3, 1, 4, 3, 4, 1, 3, 4, 1}

f[Range@4, 20, 2]

{4, 3, 2, 3, 4, 4, 3, 4, 1, 3, 3, 2, 4, 2, 3, 4, 1, 2, 3, 1}