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edit (2): Rather than using results about diffeomorphism groups of 3-manifolds you could approach the problem of showing the maps $SO_2 \to Emb(S^1, \mathbb{R^3})$ are non-trivial in a more differential-geometric way. In my "A family of embedding spaces" I give a rather direct argument that the space of self-embeddings of $\mathbb R \times D^2$ with support contained in $I \times D^2$ has the homotopy-type of $\mathbb{Z} \times K_{3,1}$, i.e. this is describing the difference between knots with trivializations of their tubular neighbourhoods vs. just "plain knots". You can soup-up this argument to show that the entire space of embeddings $Emb(S^1, \mathbb{R^3})$ can be inflated to embeddings that come equipped with trivializations of their tubular neighbourhoods with two conditions (a) the framing is the "homological framing" and (b) the framings are taken up to a coherent normal rotation by elements of $SO_2$. So from this point of view, given an element in $\pi_1 Emb(S^1, \mathbb{R^3})$ one can evaluate the embedding at the image of $\{1\} \subset S^1$ together with its framing, i.e. giving a well-defined map $\pi_1 Emb(S^1, \mathbb{R^3}) \to \pi_1 SO_3 \equiv \mathbb{Z}_2$. This argument only uses the homotopy classification of tubular neighbourhoods.

Yes, the entire homotopy-type has been computed in a sense. Although the homotopy-type is described via a chain of fiber-bundles.

The space $Emb(S^1, \mathbb{R^3})$ has the homotopy-type of a bundle

$$ (C \rtimes K_{3,1}) \times_{SO_2} SO_3$$

where $K_{3,1}$ is the space of embeddings of $I$ into $D^3$ that agree with a fixed linear embedding on the end points. $K_{3,1}$ is sometimes called the space of long-knots. The action of $SO_2$ is the rotations in the plane orthogonal to the linear axis.

$C \rtimes K_{3,1}$ is the tautological knot-exterior bundle over $K_{3,1}$ i.e. it consists of pairs of points $(p,f)$ with $f \in K_{3,1}$ and $p$ in the exterior $D^3 \setminus img(f)$.

The above result appears in my "Family of embedding spaces" paper, proposition 2.2 (on the arXiv version). The map giving the homotopy-equivalence is quite natural. You extend elements of $K_{3,1}$ to be embeddings $\mathbb{R} \to \mathbb{R^3}$, extending by the linear function, then apply stereographic projection to convert it to an embedding $S^1 \to S^3$ that agrees with a fixed linear-embedding on a prescribed arc. The point in $C$ inherits the standard framing from $\mathbb R^3$, and we project that framed point to be in the knot exterior in $S^3$. Now you do stereographic projection from that framed point to get an embedding $S^1 \to \mathbb{R^3}$, i.e. think of the framed point as an element of $SO_4$. The $SO_2$ action comes from the fact that one can rotate the knot, or use the rotations of $\mathbb R^3$ to get the same knot.

The homotopy-type of $K_{3,1}$ has been the subject of a few papers. A brief description appears in the "Family..." paper. The most compact description I know of, using the language of operads, appears in "An operad for splicing" and a longer but more basic description appears in "Topology of spaces of knots in dimension 3". Both can be found on the arXiv. The components of $K_{3,1}$ have the homotopy-type of compact manifolds. All the components are finitely-covered by a product of configuration spaces in the plane, but often the cover needs to be quite big. These manifolds include things like Klein bottles, more generally a class of Bieberbach manifolds, plus configuration spaces in the plane and various twisted products.

There is a similar description of the homotopy-type of $Emb(S^1, S^3)$ in the "Family..." paper.

Yes, the entire homotopy-type has been computed in a sense. Although the homotopy-type is described via a chain of fiber-bundles.

The space $Emb(S^1, \mathbb{R^3})$ has the homotopy-type of a bundle

$$ (C \rtimes K_{3,1}) \times_{SO_2} SO_3$$

where $K_{3,1}$ is the space of embeddings of $I$ into $D^3$ that agree with a fixed linear embedding on the end points. $K_{3,1}$ is sometimes called the space of long-knots. The action of $SO_2$ is the rotations in the plane orthogonal to the linear axis.

$C \rtimes K_{3,1}$ is the tautological knot-exterior bundle over $K_{3,1}$ i.e. it consists of pairs of points $(p,f)$ with $f \in K_{3,1}$ and $p$ in the exterior $D^3 \setminus img(f)$.

The above result appears in my "Family of embedding spaces" paper, proposition 2.2 (on the arXiv version). The map giving the homotopy-equivalence is quite natural. You extend elements of $K_{3,1}$ to be embeddings $\mathbb{R} \to \mathbb{R^3}$, extending by the linear function, then apply stereographic projection to convert it to an embedding $S^1 \to S^3$ that agrees with a fixed linear-embedding on a prescribed arc. The point in $C$ inherits the standard framing from $\mathbb R^3$, and we project that framed point to be in the knot exterior in $S^3$. Now you do stereographic projection from that framed point to get an embedding $S^1 \to \mathbb{R^3}$, i.e. think of the framed point as an element of $SO_4$. The $SO_2$ action comes from the fact that one can rotate the knot, or use the rotations of $\mathbb R^3$ to get the same knot.

The homotopy-type of $K_{3,1}$ has been the subject of a few papers. A brief description appears in the "Family..." paper. The most compact description I know of, using the language of operads, appears in "An operad for splicing" and a longer but more basic description appears in "Topology of spaces of knots in dimension 3". Both can be found on the arXiv. The components of $K_{3,1}$ have the homotopy-type of compact manifolds. All the components are finitely-covered by a product of configuration spaces in the plane, but often the cover needs to be quite big. These manifolds include things like Klein bottles, more generally a class of Bieberbach manifolds, plus configuration spaces in the plane and various twisted products.

There is a similar description of the homotopy-type of $Emb(S^1, S^3)$ in the "Family..." paper.

edit:

To answer the new question in your edit, yes all path components of $Emb(S^1, \mathbb{R^3})$ have non-trivial fundamental group, but there is likely a more elementary argument to the same effect. i.e. I think for just this result you do not need all the above machinery. Let me give a sketch.

$Emb(S^1, \mathbb{R^3})$ has an action of $SO_2$ coming from reparametrizing the embedding, i.e. pre-composing with a diffeomorphism of the domain $S^1$. So for all path components of $Emb(S^1, \mathbb{R^3})$ you have a map

$$SO_2 \to Emb(S^1, \mathbb{R^3})$$

given by re-parametrizing a representative.

I think you should be able to prove this loop is non-trivial for all knot types, since the induced diffeomorphism of the knot exterior (in $\mathbb{R^3}$) is a type of generalized Dehn twist about the peripheral torus, i.e. one can consider the action of the generator of $\pi_1 SO_2$ on $\pi_0 Diff(\mathbb{R^3} \setminus K)$, and think of this as a once-punctured knot exterior in $S^3$. The argument has a special case for the unknot vs. other knots, having to do with comparing the homotopy type of framed vs. unframed knots. But that's the rough idea.

Yes, the entire homotopy-type has been computed in a sense. Although the homotopy-type is described via a chain of fiber-bundles.

The space $Emb(S^1, \mathbb{R^3})$ has the homotopy-type of a bundle

$$ (C \rtimes K_{3,1}) \times_{SO_2} SO_3$$

where $K_{3,1}$ is the space of embeddings of $I$ into $D^3$ that agree with a fixed linear embedding on the end points. $K_{3,1}$ is sometimes called the space of long-knots. The action of $SO_2$ is the rotations in the plane orthogonal to the linear axis.

$C \rtimes K_{3,1}$ is the tautological knot-exterior bundle over $K_{3,1}$ i.e. it consists of pairs of points $(p,f)$ with $f \in K_{3,1}$ and $p$ in the exterior $D^3 \setminus img(f)$.

The above result appears in my "Family of embedding spaces" paper, proposition 2.2 (on the arXiv version). The map giving the homotopy-equivalence is quite natural. You extend elements of $K_{3,1}$ to be embeddings $\mathbb{R} \to \mathbb{R^3}$, extending by the linear function, then apply stereographic projection to convert it to an embedding $S^1 \to S^3$ that agrees with a fixed linear-embedding on a prescribed arc. The point in $C$ inherits the standard framing from $\mathbb R^3$, and we project that framed point to be in the knot exterior in $S^3$. Now you do stereographic projection from that framed point to get an embedding $S^1 \to \mathbb{R^3}$. The $SO_2$ action comes from the fact that one can rotate the knot, or use the rotations of $\mathbb R^3$ to get the same knot.

The homotopy-type of $K_{3,1}$ has been the subject of a few papers. A brief description appears in the "Family..." paper. The most compact description I know of, using the language of operads, appears in "An operad for splicing" and a longer but more basic description appears in "Topology of spaces of knots in dimension 3". Both can be found on the arXiv. The components of $K_{3,1}$ have the homotopy-type of compact manifolds. All the components are finitely-covered by a product of configuration spaces in the plane, but often the cover needs to be quite big. These manifolds include things like Klein bottles, more generally a class of Bieberbach manifolds, plus configuration spaces in the plane and various twisted products.

There is a similar description of the homotopy-type of $Emb(S^1, S^3)$ in the "Family..." paper.

Yes, the entire homotopy-type has been computed in a sense. Although the homotopy-type is described via a chain of fiber-bundles.

The space $Emb(S^1, \mathbb{R^3})$ has the homotopy-type of a bundle

$$ (C \rtimes K_{3,1}) \times_{SO_2} SO_3$$

where $K_{3,1}$ is the space of embeddings of $I$ into $D^3$ that agree with a fixed linear embedding on the end points. $K_{3,1}$ is sometimes called the space of long-knots. The action of $SO_2$ is the rotations in the plane orthogonal to the linear axis.

$C \rtimes K_{3,1}$ is the tautological knot-exterior bundle over $K_{3,1}$ i.e. it consists of pairs of points $(p,f)$ with $f \in K_{3,1}$ and $p$ in the exterior $D^3 \setminus img(f)$.

The above result appears in my "Family of embedding spaces" paper, proposition 2.2 (on the arXiv version). The map giving the homotopy-equivalence is quite natural. You extend elements of $K_{3,1}$ to be embeddings $\mathbb{R} \to \mathbb{R^3}$, extending by the linear function, then apply stereographic projection to convert it to an embedding $S^1 \to S^3$ that agrees with a fixed linear-embedding on a prescribed arc. The point in $C$ inherits the standard framing from $\mathbb R^3$, and we project that framed point to be in the knot exterior in $S^3$. Now you do stereographic projection from that framed point to get an embedding $S^1 \to \mathbb{R^3}$, i.e. think of the framed point as an element of $SO_4$. The $SO_2$ action comes from the fact that one can rotate the knot, or use the rotations of $\mathbb R^3$ to get the same knot.

The homotopy-type of $K_{3,1}$ has been the subject of a few papers. A brief description appears in the "Family..." paper. The most compact description I know of, using the language of operads, appears in "An operad for splicing" and a longer but more basic description appears in "Topology of spaces of knots in dimension 3". Both can be found on the arXiv. The components of $K_{3,1}$ have the homotopy-type of compact manifolds. All the components are finitely-covered by a product of configuration spaces in the plane, but often the cover needs to be quite big. These manifolds include things like Klein bottles, more generally a class of Bieberbach manifolds, plus configuration spaces in the plane and various twisted products.

There is a similar description of the homotopy-type of $Emb(S^1, S^3)$ in the "Family..." paper.