Let me elaborate a bit on my comments.

First of all, algebraic knots (up to **isotopy**) have been classified. People refer to an unpublished paper by Bonahon and Siebenmann, *The classification of algebraic links*, and there appears to be a discussion of this fact in the book *Three-dimensional link theory and invariants of plane curve singularities* by Eisenbud and Neumann (ed. Princeton University Press): they're some iterated cables of the unknot (some positivity condition is involved, as -for example- negative torus knots are not algebraic).

**1**. Your question, on the other hand, seems to be more focused on a given embedding. Let's assume that your definition of algebraic is that you identify $S^3_\varepsilon$ with $S^3_1$ using the rescaling in $\mathbb{C}^2$, and define $K$ to be the embedding of $K\simeq S^1$ in $S^3$ using these identifications*. My guess would be that, using this definition, most smooth perturbations of $K$ are *not* algebraic (mostly because real functions are many more than complex ones).

On the other hand, it's easy to produce a non-algebraic knot (under the same definition) in any isotopy class of knots; call $J$ the canonical complex structure on $\mathbb{C}^2$: let's observe that if $K$ is algebraic, parametrised by $\gamma$, then $J(\dot\gamma)$ lies in $T(f^{-1}(0))$ and is linearly independent of $\dot\gamma$, therefore it can't lie in $TS^3$. Therefore, $\dot\gamma$ is never a complex tangency to $S^3$ (*i.e.* $\dot\gamma \not\in TS^3\cap JTS^3$, that is, $K$ is a transverse knot with respect to the standard contact structure on $S^3$). In particular, take a given algebraic $K$, then you can just bend it so that somewhere it has a complex tangency, and that's an obstruction to being algebraic. (As it happens, you can also ask for a Legendrian approximation, but that's more than we need)

**2**. Let me now assume that you're talking about algebraic knots as isotopy classes of algebraic knots (as in the previous definition). As I said, algebraic knots have been classified, so the question is -in principle- solved. Still, there are some nice obstructions to being algebraic. As you wrote, an algebraic knot gives an open book for $S^3$, and an open book supports a contact structure: an algebraic knot has to support the standard contact structure (the one defined above, *i.e.* the complex tangencies to $S^3$). Now, $S^3$ admits (infinitely many) overtwisted contact structures (Bennequin) and every contact structure is supported by an open book, which is unique up to stabilisations (Giroux). In particular, every contact structure admits an open book with connected binding. Let's take any overtwisted $\xi$ on $S^3$, take an open book with connected binding $K$ supporting $\xi$: $K$ is fibred (by definition), but not algebraic.

The question of whether a fibred knot (hence, an open book) supports the standard contact structure has in turn been completely solved (Etnyre and Van Horn-Morris): a fibred knot $K$ supports $\xi_{\rm st}$ if and only if it's quasi-positive, and in turn this holds if and only if $g(K) = \tau(K)$ (where $\tau$ is the concordance invariant of $K$ coming from knot Floer homology), and this holds if and only if the maximal self-linking number of $K$ (that is, the maximal self-linking number of a transverse representative of $K$ with respect to $\xi_{\rm st}$) is $2g(K)-1$.

In particular, the figure-eight knot has maximal self-linking number -3 (and $\tau = 0$), so it can't be algebraic. Also, the mirror of any nontrivial algebraic knot is not algebraic (since $\tau$ changes sign when taking the mirror, but probably also using "classical" inequalities for the self-linking number coming from knot polynomials).

* This embedding is defined up to self-diffeomorfisms of $S^1$.