Fix $d\in\mathbb{N}$. Consider the following sets as topological spaces with the subspace topology from $\mathbb{R}^{d+1}$. $$S^d = \{ (x_0,\ldots,x_d)\in\mathbb{R}^{d+1}\mid \sum x_i^2 = 1\}$$ $$ D^{d+1} = \{ (x_0,\ldots,x_d)\in\mathbb{R}^{d+1}\mid \sum x_i^2 \leq 1\}.$$
Is it true that for every continuous map $f:S^d\to\mathbb{R}^d$, there is a continuous map $g:D^{d+1}\to\mathbb{R}^d$ such that $g|_{S^d}\equiv f$, and $g(D^{d+1})\subseteq f(S^d)$?
As requested, here is my comment in answer form:
This fails for $d=3$. The Hopf map $f : S^3\to S^2\subset \mathbb{R}^3$ cannot be homotoped (through maps to $S^2$) to a constant map. The existence of $g : D^4 \to \mathbb{R}^3$ with $g(D^4)\subset f(S^3) = S^2$ would precisely be such a homotopy.
For $d=2$ the question asks if a map $f:S^2 \to \mathbb{R}^2$ can always be homotoped to a trivial map inside of its image. I think the answer to this is "yes" it can: Ian Agol's answer here points to a paper of Cannon-Conner-Zastrow proving that $\pi_2(f(S^2)) = 0$. This yields the desired homotopy.
(Of course, as pointed out in the comments, for $d=1$ a continuous map $f:S^1 \to \mathbb{R}$ has an interval as an image, and thus is homotopically trivial in its image.)