YES

We show the termination of R/S, where R is

  p(s(x)) -> x
  fac(0) -> s(0)
  fac(s(x)) -> times(s(x),fac(p(s(x))))

and S is:

  rand(x) -> x
  rand(x) -> rand(s(x))

Since R almost dominates S and S is non-duplicating,
it is enough to show finiteness of (P, Q).
Here P consists of the dependency pairs

  fac#(s(x)) -> fac#(p(s(x)))
  fac#(s(x)) -> p#(s(x))

and Q consists of the rules:

  p(s(x)) -> x
  fac(0) -> s(0)
  fac(s(x)) -> times(s(x),fac(p(s(x))))
  rand(x) -> x
  rand(x) -> rand(s(x))

The weakly monotone algebra (N^3, >_lex) with

  fac#_A((x1,y1,z1)) = (1, 0, 0)
  s_A((x1,y1,z1)) = (x1, y1, z1 + 1)
  p_A((x1,y1,z1)) = (x1, y1 + 1, 2)
  p#_A((x1,y1,z1)) = (0, 0, 0)
  fac_A((x1,y1,z1)) = (x1 + 1, 1, 1)
  0_A = (1, 2, 1)
  times_A((x1,y1,z1),(x2,y2,z2)) = (x2, 0, 0)
  rand_A((x1,y1,z1)) = (x1 + 1, y1 + 1, 0)

strictly orients the following dependency pairs:

  fac#(s(x)) -> p#(s(x))

We remove them from the set of dependency pairs.

The weakly monotone algebra (N^3, >_lex) with

  fac#_A((x1,y1,z1)) = (x1, y1, z1)
  s_A((x1,y1,z1)) = (x1, y1 + 1, 2)
  p_A((x1,y1,z1)) = (x1, y1, 1)
  fac_A((x1,y1,z1)) = (x1 + 1, y1 + 2, 2)
  0_A = (1, 1, 0)
  times_A((x1,y1,z1),(x2,y2,z2)) = (x1, 4, 3)
  rand_A((x1,y1,z1)) = (x1 + 1, 1, 0)

strictly orients the following dependency pairs:

  fac#(s(x)) -> fac#(p(s(x)))

We remove them from the set of dependency pairs.

No dependency pair remains.