The result is true in general.
We may assume a counterexample is given in the form of a domain $R$ satisfying the second property but with nontrivial Jacobson radical, i.e. the closed points of Spec $R$ are not dense. Let D be an affine open neighborhood of $(0)$ in Spec $R$ which contains no closed points. Since D is affine, there exists some $x\in$ D which is closed in D. That is,
$\overline{\lbrace x\rbrace}\setminus x\subset\text{Spec }R\setminus D$.
Since $D$ is open, this implies
$x\not\in \overline{\overline{\lbrace x\rbrace}\setminus x}$,
but this contradicts the requirements of the second property.
One general fact that comes to mind: If an ideal $I\subset \mathbb{k}[x_1,\dots,x_n]$ contains an element of the form $f = gx_1 + h$ where $g,h$ don't use $x_1$, and $g$ is a nonzerodivisor mod $I$, then the primary components of $I\cap \mathbb{k}[x_2,\dots,x_n]$ and $I$ are in bijection. This is birational projection and I learned it from Mike Stillman (see Proposition 23 in the appendix of http://arxiv.org/pdf/math/0301255.pdf).
Now here is almost a counter-example to your question:
$$ I = \langle x_{1} x_{9}-x_{4}x_{8}, x_{4}x_{6}-x_{7}x_{9}, x_{2}x_{5}-x_{3}x_{9}, x_{2}x_{3}-x_{5}x_{6} \rangle \subset \mathbb{k}[x_1,\dots,x_9]$$
This ideal has 6 components, one of which is primary with minimal prime $\langle x_9, x_5, x_4, x_2 \rangle$.
If I read your hypotheses correctly, the only bit missing is the pairwise different degrees of the generators. I have an inkling that this may be a red herring. If I modify my example by adding some extra unrelated variables, then the embedded component over $\langle x_9, x_5, x_4, x_2 \rangle$ is essentially unchanged:
$$\langle x_{1}x_{9}-x_{4}x_{8}, x_{4}x_{6}y_{1}-x_{7}x_{9}y_{2}, x_{2}x_{5}y_{3}y_{4}-x_{3}x_{9}y_{5}y_{6}, x_{2}x_{3}y_{7}y_{8}y_{9}-x_{5}x_{6}y_{10}y_{11}y_{12} \rangle$$
The Binomials package in Macaulay2 quickly confirms that this ideal is not radical.
Best Answer
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 $R$