Simply connected will only ensure that closed one forms are exact. If you delete the origin from $R^3$, there should be a 2-form on this space which is closed but not exact (I think you could write it down by pulling back the volume form of the sphere to $R^3-0$).
In general closed forms will always be exact on contractible spaces.
I would recommend reading a book on de Rham cohomology, such as Bott and Tu, or From Calculus to Cohomology.
No. "Every closed form is exact" is equivalent to the claim that the first de Rham cohomology $H^1_{dR}(U, \mathbb{R})$ vanishes. This means, equivalently, that there are no nontrivial homomorphisms from the fundamental group $\pi_1(U)$ to $\mathbb{R}$. But it does not imply that the fundamental group is trivial (which is equivalent to simply connected).
In fact any reasonable space (e.g. a manifold) is homotopy equivalent to an open subset of some $\mathbb{R}^n$ (embed it into $\mathbb{R}^n$ in a nice way, then take a tubular neighborhood), and lots of reasonable spaces have the property that their fundamental group is nontrivial but admits no nontrivial maps to $\mathbb{R}$. Maybe the simplest such space is the real projective plane $\mathbb{RP}^2$, which embeds nicely into $\mathbb{R}^4$ (I think) and has fundamental group $\mathbb{Z}_2$.
If you really want an explicit open subset of some $\mathbb{R}^n$, you can take $GL_3^{+}(\mathbb{R})$, the space of $3 \times 3$ matrices with positive determinant, which is homotopy equivalent to $SO(3)$, which is in turn the real projective space $\mathbb{RP}^3$, and which in particular again has fundamental group $\mathbb{Z}_2$. This is a connected open subset of $\mathbb{R}^9$.
In general, homomorphisms $\pi_1(U) \to \mathbb{R}$ correspond to homomorphisms $H_1(U, \mathbb{Z}) \to \mathbb{R}$, where $H_1$ is the first singular homology. If the fundamental group $\pi_1(U)$ is finitely generated (which is again true in all reasonable cases), every such homomorphism is trivial iff $H_1(U, \mathbb{Z})$ is torsion iff it is finite.
Best Answer
Let $U=\mathbb R^n\setminus \lbrace0 \rbrace\subset \mathbb R^n$, a simply connected domain for $n\geq3$, which we assume from now on.
The $(n-1)$ form $\omega \in \Omega^{n-1}(U)$ defined by
$$\omega (x)=\frac {1}{\mid \mid x\mid \mid^n} \sum _{i=1} ^n (-1)^{i-1}x_idx^1\wedge...\wedge \widehat{dx^i} \wedge dx_n$$ for $x\in \mathbb R^n\setminus \lbrace0 \rbrace)$ is closed but not exact. It is thus an example of what you want.
More precisely, its cohomology class $[\omega ]$ generates the one-dimensional $(n-1)$-th De Rham cohomology vector space of $U$, namely $H^{n-1}_{DR}(U)=\mathbb R\cdot[\omega ]$
NB
a) This form can also be seen as the pull-back $\omega =r^*(vol)$ of the canonical volume form $vol\in \Omega ^{n-1}(S^{n-1})$ of the unit sphere under the map $$r:U\to S^{n-1}:x\mapsto \frac {x}{\mid \mid x\mid \mid}$$
b) To be quite explicit, the value of the alternating form $\omega (x)$ on the $(n-1)$-tuple of vectors $v_1,...,v_{n-1}\in T_x(U)=\mathbb R^n$ is $$\omega (x)(v_1,...,v_{n-1})=\frac {1}{\mid \mid x\mid \mid^n}\cdot det[x\mid v_1\mid ...\mid v_{n-1}]$$
where of course $x, v_1, ...,v_{n-1}$ are seen as column vectors of size $n$.