Is this argument wrong "since DOM is special kind of RDOM, then RDOM is NP-hard"?
There are two statements here that rule the argument. The first statement:


  Statement 1.
  To show that $RDOM$ is $\mathbf{NP}$-hard, one must show that $DOM
 \leq_P RDOM$, i.e. given an instance $\langle G,k \rangle$, construct an instance
  $\langle G',r,k'\rangle$ in polynomial time such that $\langle G,k \rangle \in DOM$ if and only
  if $\langle G',r,K'\rangle \in RDOM$.


is a general framework for showing that RDOM is NP-hard. However, in this statement, you have not shown how to prove that this reduction works. You just said how to prove that this problem (or any problem) is NP-hard. In order to prove it, you need a proper reduction from all instances of DOM to instances of RDOM. The second argument however,


  Statement 2.
  By the way, my friend states that we know that a domination is a
  $1$-rainbow domination in the same graph, i.e. domination is a special
  kind of rainbow-domination, and since domination is
  $\mathbf{NP}$-hard, then $RDOM$ is also $\mathbf{NP}$-hard. How to
  convince him that this is wrong? or Is this argument correct?


looks like a proper reduction. If "every domination is a $1$-rainbow domination in the same graph", then this proof is completely right. This is a proper reduction from every instance of DOM problem to instances of RDOM or 1DOM problem. However, if "every domination is NOT a $1$-rainbow domination in the same graph", then this proof is completely wrong.

My understanding is that a domination, can be any of 1-rainbow domination, ...,or |V|-rainbow domination. In this case, it is implied that by having a polynomial algorithm for RDOM problem, you can iterate $|V|$ times thorough $r$-dominations and determine if the graph is dominating. Thus, if you have a polynomial algorithm for RDOM problem, you can solve the DOM problem polynomially.