Is every knot unavoidable in the embeddings of some graph? 
Is it the case that, for any given knot $K$,
  there exists some graph $G$ whose every embedding into $\mathbb{R}^3$
  (or into $\mathbb{S}^3$)
  contains a cycle that realizes $K$?

I know the famous Conway-Gordon result that the complete graph
on seven vertices $K_7$ is intrinsically knotted in that every
embedding contains a knotted cycle.
And Joel Foisy proved that $K_{3,3,1,1}$ is also intrinsically knotted
(J. Graph Theory, 2002, ACM link).
Instead, I am asking a Ramsey-like question: 
Whether it's known that, for every particular knot, there exists some
graph for which that particular knot is unavoidable.
 A: The answer is no for the following silly reason.  Take any graph $G$ and some embedding into $\mathbb R^3$.  To any edge, we may perform the connected sum with a fixed knot $K'$ (that is, take a small straight part of the edge and connect sum to it $K'$ there).  Do this operation separately to all of the edges.  Then any cycle in $G$ has knot type of the form $K'\#(\text{something})$.  We just need to pick a knot $K'$ so that the given $K$ is not of this form.
We could modify the question and ask whether a fixed knot $K$ exists in some weaker sense in any given embedding of $G$.  One such (very) weak sense could be that there exists a knotted torus $T$ in knot type $K$ and a cycle in $G$ which is contained in $T$ and cannot be made disjoint from a compressing disk for $T$.  I suspect this may also have a negative answer, however.  It seems one can probably always find embeddings of $G$ which are arbitrarily complicated, and in particular which are not related in any way to the given knot $K$.
A: Yes. See this paper of Negami.  The main result is that for any fixed knot (or link) of type $k$, there is a constant $R(k)$ such that every straight line embedding of $K_{R(k)}$ in $\mathbb{R}^3$ contains a knot (or link) equivalent to $k$.   
