$\newcommand\R{\mathbb{R}} \newcommand\cH{\mathcal{H}} \newcommand\bbD{\mathbb{D}}$ I'm marking this community wiki since it's not so much an answer but an attempt to make it clear to the casual reader that despite Nick Mendler's answer it really is obvious that for a finite set $S \subset \R^3$, a continuous curve $\gamma\colon S^1 \rightarrow \R^3 \setminus S$, and a point $y \in \R^3 \setminus S$ the set of homotopies from $\gamma$ to the constant map $y$ that avoid $S$ is open and dense in the set of all homotopies. Here "obvious" means "immediate from standard tools/arguments that have little to do with $\R^3$". All I will use about $\R^3$ is that it is a smooth 1-connected manifold of dimension at least $3$.

The most convenient way to think of such homotopies is as the set $\cH$ of all continuous maps $f\colon \bbD^2 \rightarrow \R^3$ with $f|_{S^1} = \gamma$ and $f(0) = y$. Let $\cH_S$ be the subspace of $\cH$ consisting of such $f$ whose image is disjoint from $S$. Give $\cH$ the compact-open topology. We want to prove that $\cH_S$ is open and dense in $\cH$. That it is open is genuinely obvious, so the key thing to prove is that it is dense.

Since $\mathbb{R}^3$ is $1$-connected, the set $\cH$ is nonempty. Consider some $f_0 \in \cH$. Our goal is to perturb $f_0$ slightly (I'm not going to bother keeping track of $\epsilon$'s) to move it into $\cH_S$. Choose three nested proper open annuli $A_1 \subset A_2 \subset A_3$ in $\bbD^2$ with $\overline{A}_1 \subset A_2$ and $\overline{A}_2 \subset A_3$ such that $f_0(\bbD^2 \setminus A_1)$ is disjoint from $S$. Here "proper open annuli" means that $0 \notin A_1$ and $S^1 \cap \overline{A}_3 = \emptyset$. That we can do this is clear: just make $A_1$ nearly all of $\bbD^2$.

The first standard fact we appeal to is that we can perturb $f_0$ to some $f_1 \in \cH$ with the following properties:

1. $f_1$ is smooth on $\overline{A}_2$; and
2. $f_1 = f_0$ on $\bbD^2 \setminus A_3$, which implies that $f_1 \in \cH$.

By making this perturbation small enough, we can assume that it is still the case that $f_1(\bbD^2 \setminus A_1)$ is disjoint from $S$.

The second standard fact we appeal to is that we can perturb $f_1$ to some $f_2 \in \cH$ with the following properties:

1. $f_2$ is smooth on $\overline{A}_2$ and is transverse to $S$ on $A_1$; and
2. $f_2 = f_1$ on $\bbD^2 \setminus A_2$, which implies that $f_2 \in \cH$.

Again, by making this perturbation small enough, we can assume that it is still the case that $f_2(\bbD^2 \setminus A_1)$ is disjoint from $S$. Since on the $2$-manifold $A_1$ the map $f$ is transverse to the $0$-submanifold $S$ of the $3$-manifold $\R^3$, we actually have that $f_2(A_1)$ is disjoint from $S$. We conclude that $f \in \cH_S$, as desired.