Yes, forcing can add fundamentally new knots, not equivalent to any ground model knot. Indeed, whenever you extend the set-theoretic universe to add new reals, then you must also have added fundamentally new knots.
Theorem. If $V\subset W$ are two models of set theory with the same ordinals and different reals (for example, any forcing extension with new reals), then there are new knots in $W$ not equivalent in $W$ to any ground model knot from $V$.
(Note that indeed it makes sense to consider the ground model knots $k$ as knots in any extension; simply view them as a Borel set, to be re-interpreted as the corresponding Borel set in the extension, constructed according to the same process; the assertion that a given Borel code represents a knot is a $\Sigma^1_1$ property and therefore absolute. One should not think of a ground model knot merely as a point set, since this exact set will no longer be a knot in a forcing extension with new real numbers.)
To prove the theorem, we use the work of Vadim Kulikov, who has already done the heavy knot-tying for this question. Specifically, in his answer to the question on the Descriptive set-theoretic complexity of knot equivalence, Vadim explains that the relation of knot equivalence is strictly above the isomorphism relation of countable structures in the hierarchy of Borel reducibility. This means that there is a Borel map $\pi$ from the space of countable structures to knots, such that structures $A$ and $B$ are isomorphic if and only if knots $\pi(A)$ and $\pi(B)$ are equivalent (but no such map in the other direction).
One should think of the map $\pi$ as a means to code an arbitrary countable structure into a knot. Using $\pi$, one can code a countable graph, group or partial order $A$ into the knot $\pi(A)$, in a way that respects knot equivalence. In this way, Vadim's result fulfills the suggestion I had made in the comments on the question that what we want to do is code arbitrary binary sequence into knots. Given the countable structure $A$, the knot $\pi(A)$ codes $A$ in that we can recover $A$ up to isomorphism from any knot that is equivalent to $\pi(A)$. The recovery process is sufficiently explicit that it is absolute between a model of set theory and its forcing extensions.
This analysis of the complexity of not equivalence is evidently the main result of his paper:
Kulikov states that it follows from his theorem that, "wild knots cannot be completely classified by real numbers considered up to any Borel equivalence relation." So there can be no classification of the wild knots along the same lines as the tame knot classification.
My point here is that it follows from the positive part of the reduction — that isomorphism of countable structures reduces to knot equivalence — that in any extension $W$ with new real numbers, there will also be new knots, not equivalent to any ground model knot.
To see this, suppose that $z$ is a real in an extension $W$, but not in $V$. Let $A_z$ be a countable structure, say, a graph, coding $z$ in some fundamental way, making it definable from the theory of this structure. Let $k=\pi(A_z)$ be the associated knot in $W$. I claim that $k$ is not equivalent to any ground model knot. If $k$ were equivalent to some $k'$ in the ground model $V$, then the assertion $\exists y\ k'\sim\pi(A_y)$ is true about $k'$ in $W$. This is a $\Sigma^1_1$ statement about $k'$, which by Lévy absolutelenss must already be true in the ground model. So there is some ground model real $y$ such that $k'$ is equivalent to $\pi(A_y)$ in $V$. Thus, $\pi(A_y)$ and $\pi(A_z)$ are equivalent in $W$, and so $A_y$ must be isomorphic to $A_z$, contradicting that $z$ is not in $V$ and therefore not definable in $A_y$.
So $k$ is a totally new knot, as desired.
This phenomenon is not possible with tame knots. Since there are only countably many tame knots up to equivalence, there is a list of countably many tame knots, and the assertion that every tame knot is equivalent to one of them is a $\Pi^1_2$ assertion about that list. By absoluteness, this remains true in all forcing extensions. So you cannot add fundamentally new tame knots.
(This answer addresses what I take to be the main part of your questions, whether forcing adds new knots, and not specifically about the crossing number issue.)
Best Answer
If you allow yourself to consider knotted trivalent graphs, instead of just knots, then you can start with just the tetrahedron and Mobius bands, and use the operations unzip, bubbling, and connect sum to get all knotted trivalent graphs. This result is due to Dylan Thurston and Dror Bar-Natan and is written up by Dylan in a GT monograph. Their motivation was the lack of a good candidate for an answer to your question which stayed entirely in the language of knots. However for many situations KTGs are just as good as knots (e.g. click on a random link at Dror's wiki).