I construct a counterexample for your question in the non-noetherian case:

**(1)** Let $A = k[[X,Y]]/(XY) = k[[x,y]]$, where $k$ be a field. Notice that $(0) = (x) \cap (y)$ so $(0)$ is reducible in $A$.

**(2)** We consider the injective hull $E(k)$ of $k$, and set $m \in E(k)$ be the element such that $mA \cong k$. Notice that every non-zero submodule of $E(k)$ contains $mA$ and $(0)$ is irreducible in $E(k)$

**(3)** Set $R = A \ltimes E(k)$ be the indealization. We have that $(0 \ltimes E(k))^2 = 0$ so $\sqrt{(0)R} = 0 \ltimes E(k)$ is reducible by (1).

**(4)** We can prove that for every non-zero element $(a,s)$ of $R$, we have $0 \ltimes mA \subseteq (a,s)R$. So the ideal $(0)$ is irreducible in $R$.


EDIT (13/02): It should be noted that this example is also a counterexample for a non-noetherian ring with an ideal is irreducible but not primary. Indeed, we have $(0)$ is irreducible as above. However
$$(x,0).(y,m) = (0,0) \in R,$$
and $(x,0)$ and $(y,m)$ are not nilpotent so $(0)$ is not primary in $R$