I prove your question for (not necessarily Noetherian) commutative ring. Irreducible ideals in non-Noetherian ring are complicate (see [this question][1]). For Noetherian ring, see [our paper][2] for a study of the index of reducibility. The proof is elementary but quite long.

**Edit:** I realise that the proof also work for [another question][3]. If$(0)$ is an irreducible ideal of $R$ then it is a graded irreducible ideal of the graded ring $R[X]$. 



We can assume that $I = 0$. Suppose $ 0 \in R[X]$ is reducible as the intersection of two proper ideals. We can assume these two ideals are principal, so $0 = (f) \cap (g)$ with
$$f = X^r (a_0 + a_1X + \cdots + a_nX^n), a_0 \neq 0,$$
$$g = X^s (b_0 + b_1X + \cdots + b_mX^m), b_0 \neq 0.$$
Choose $f$ and $g$ so that $m+n$ is **minimal**.

**Fact:** Let $f = X^rf'$ and $g = X^sg'$. Then $(f) \cap (g) = 0$ if and only if $(f') \cap (g') = 0$.

**Proof.** The "if part" is clear since $(f) \subseteq (f')$ and $(g) \subseteq (g')$.
For the "only if" part, suppose $(f') \cap (g') \neq 0$. We have $(X^{r+s} f') \cap (X^{r+s}g') = X^{r+s}((f') \cap (g')) \neq 0$. Thus $(f)\cap (g) \neq 0$ since $(X^{r+s}f') \subseteq (f)$ and $(X^{r+s}g') \subseteq (g)$. 
 
Using the above fact we can assume 
$$f = a_0 + a_1X + \cdots + a_nX^n, a_0 \neq 0$$
$$g = b_0 + b_1X + \cdots + b_mX^m, b_0 \neq 0.$$

If $m+n = 0$ then $f, g \in R$ and this contradicts the assumption that $(0)$ is irreducible there. 

So $m+n> 0$.  Without loss of generality we assume that $m \ge n$ and thus $m>0$. Choose $0 \neq c \in (a_0) \cap (b_0)$ in $R$.  Then $c = da_0 = eb_0$ for some $d, e \in R$. Replacing $f$ and $g$ by $df$ and $eg$, respectively, we can assume henceforth that $a_0 = b_0$.

By the minimality of $m+n$ we have the following.

**Claim 1:** Let $r$ be an element of $R$ such that $ra_0 = 0$. Then $rf = 0$ and $rg = 0$.

**Proof.** Assume $ra_0 = 0$ and $rf \neq 0$. Since $(f) \cap (g) = 0$ also $(rf) \cap (g) = 0$. However, since $ra_0 = 0$, $rf = X^tf'$ with $\mathrm{deg}(f')<n$. By the Fact, $(f') \cap (g) = 0$ in contradiction to  minimality of $m+n$.  The same argument applies to $g$.

Applying Claim 1 inductively one can prove:

**Claim 2:** If a polynomial $h = c_0 + c_1X + \cdots + c_k X^k$ satisfies $hf = 0$ (resp. $hg = 0$), then each coefficient $c_i$ satisfies $c_if = 0$ (resp. $c_ig = 0$).

Combining Claims 1 and 2 we have

**Claim 3:** $hf = 0$ if and only if $hg = 0$.


We continue the proof. Let $g' = g-f$. Since $a_0 = b_0$, we get $g' = X^{m-m'}g' '$ for some $m' < m$ and polynomial $g' '$ of degree $m'$.
By minimality of $m+n$ and the Fact, we have $(f) \cap (g' ') \neq 0$. Thus there are polynomials $u,v,w$ such that $0 \neq w = uf = vg' '$, and thus $X^{m-m'}uf = vg' = v(g-f)$.  Now, if $vg \neq 0$ then $vg = (u+v)f \in (f) \cap (g)$, a contradiction. Therefore $vg = 0$. By Claim 3 we have $vf = 0$ so $w = 0$. This is also a contradiction. The proof is complete.


  [1]: http://mathoverflow.net/questions/87870/is-the-radical-of-an-irreducible-ideal-irreducible/88215#88215
  [2]: http://arxiv.org/abs/1405.1136
  [3]: http://mathoverflow.net/questions/162739/graded-irreducible-ideals-are-irreducible/216106#216106