I need to show the statement in the title, and would like to have some checking on the argument below.
Let $f: S^1 \to X$ be a map, $p \in X$, $e_p: S^1 \to X$ be the constant map, i.e. $e_p(z) = p,\forall z \in S^1$.
The condition $f \simeq e_p$ implies the existence of a map $H: S^1 \times I \to X$ such that $H(z,0) = f(z)$, $H(z,1) = e_p(z) ,\forall z \in S^1$. My intuition tells me that, because of the way $H(z,1)$ is defined, the domain of this map can be re-written as $$ (S^1 \times I)/(S^1 \times \{1\}), $$ which is homeomorphic to $D^2$. So this modified homotopy is the extension we're looking for. But I'm not sure how to make this argument precise and which conditions that I need to check.
Consider the map $q$ from $S^1 \times [0, 1]$ to $D^2$ given by
$$ (\theta, t) \mapsto ((1-t) \cos \theta, (1-t) \sin \theta) $$
This is constant on $S^1 \times \{1\}$, and in fact a homeomorphism from $S^1 \times [0, 1] / S^1 \times \{1\}$ to $D^2$, so if we have a function, $$ H : S^1 \times [0, 1] \to X $$ that's constant on $S^1 \times \{1\}$, then $$ H' = H \circ q^{-1} $$ is a continuous map $D^2 \to X$.
Now suppose that $s : S^1 \to X$ is given. By hypothesis, there's a homotopy $$ H: S^1 \times [0, 1] \to X $$ with the property that $H(\theta, 1)$ is constant. Then define $H'$ as above to get a map from $D$ to $X$ that extends $s$.