Is a knot in $\mathbb{R}^3$ that can be untied necessarily trivial? Is a knot in  $\mathbb{R}^3$ that can be untied necessarily trivial?
Trivial means it is equivalent(of the same knot type) to a circle in a plane.
A knot K is the image of a homeomorphism  of the unit circle into $\mathbb{R}^3$.
K is said to be trivial if there is a homeomorphism of $\mathbb{R}^3$ to $\mathbb{R}^3$ such that K is the image of its restriction on the unit circle.
I don't know exactly how untie is defined. 
An example from book  Introduction to KNOT THEORY by R H Fox is 

The number of loops increases without limit while their size decreases without limit when approaching point $p$. 
It can be untied but it is wild( means  that is not of knot type of a  polygonal knot). But a unit circle is obviously not wild.
So what is the difference between the two notions?
 A: Intuitive explanation is the following: start untying the knot by pulling straight the rightmost knotted strand and then continue further to the left by repeating the same move. The single move is obviously an isotopy, but you would have to repeat it infinitely many times as you approach $p$ in order to get to the unknot. And with such infinite procedures, one has to be careful.
Careful and precise description is through the following notions (as in the paper by R.H.Fox "A Remarkable Simple Closed Curve"):


*

*Two knots are equivalent if there is an orientation-preserving self-homeomorphism of $\mathbb{R}^3$ which transforms one into another.

*A knot $K$ is almost unknotted if for any point $p$ of $K$ and any neighbourhood $U\subset \mathbb{R}^3$ of $p$ there exists a neighbourhood $V\subset U$ of $p$ and a homeomorphism $\varphi$ of $\mathbb{R}^3$ such that $\varphi|_{V} = id_V$, and $\varphi(K \setminus V)$ is contained in a plane.
The knot in the figure is an example of an almost unknotted knot that is not equivalent to the unknot. Namely, for a small neighbourhood $V$ of the distinguished point $p$, we can achieve that part of the knot which is not in $V$ lies in a plane by performing finitely many untying moves (so through a homeomorphism of $\mathbb{R}^3$). 
Fox proved that the fundamental group of the complement of this knot is non-abelian, so it cannot be equivalent to the unknot.
