Consider an embedding of $\mathbb R$ into $\mathbb R^3$ which consits of an infinite connected sum of knots. There is a path in the space of proper embeddings $\mathbb R\hookrightarrow\mathbb R^3$ which goes from that knotted line to the unknotted embedding of $\mathbb R$ into $\mathbb R^3$ (where $\mathbb R$ maps to the $x$-axis). That's the path that "sends all the knotted junk to infinity".

Now, there is no path in $Diff(\mathbb R^3)$ that lifts that path in $Emb(\mathbb R,\mathbb R^3)$. That's because the isn't any diffeomorphism of $\mathbb R^3$ that maps the the $x$-axis to that knotted curve (the proof uses the fundamental group at infinity of $\mathbb R^3$ minus the image of $\mathbb R$).

This shows that your map is not even a Serre fibration.

Consider an embedding of $\mathbb R$ into $\mathbb R^3$ which consits of an infinite connected sum of knots, arranged roughly along the $x$-axis. There is a path in the space of proper embeddings $\mathbb R\hookrightarrow\mathbb R^3$ which goes from that knotted line to the unknotted embedding of $\mathbb R$ into $\mathbb R^3$ (where $\mathbb R$ just maps to the $x$-axis). That's the path that "sends all the knotted junk to infinity".

Now, there is no path in $Diff(\mathbb R^3)$ that lifts that path in $Emb(\mathbb R,\mathbb R^3)$. That's because the isn't any diffeomorphism of $\mathbb R^3$ that maps the the $x$-axis to that knotted curve (the proof uses the fundamental group at infinity of $\mathbb R^3$ minus the image of $\mathbb R$).

This shows that your map is not even a Serre fibration.