Show that $G$ is a path homotopy between $\gamma$ and the constant loop $e_C$.
This should imply that $[\gamma] = [e_C]$ which means $\ker (p_*)$ is trivial, which means that $p_*$ is injective.
If $f:\mathbb{S}^1 \to \mathbb{S}^1$ were homotopic to a constant map, you would need the homotopy to stay within $\mathbb{S}^1$, $$F:\mathbb{S}^1\times I \to \mathbb{S}^1$$ But it doesn't: $$|(1-t)f(\theta)|=|1-t||f(\theta)|=|1-t|\neq 1, \text{ for every } t\neq 0.$$
So you have the identity on the disk homotopic to a constant (what you have proven via $F:D\times I \to D$), but the identity on the sphere cannot be homotopic to a constant via your homotopy via restriction of your homotopy (because $F(\mathbb{S}^1\times I)\nsubseteq \mathbb{S}^1$, as we have shown above).
EDIT: It is true that $F:D\times I \to D$ given by $F(x,t)=tx_0+(1-t)x$ is a straightline homotopy between the identity function $id:D\to D$ and the constant any point $x_0$ of the disk, $c:D\to D$, $c: x\mapsto x_0$. Just check the definition for this homotopy.
Your confusion arises because the degree of a map is defined for maps $f:\mathbb{S}^n\to \mathbb{S}^n$. It is precisely the integer $d$ such that $H_n(f):H_n(\mathbb{S}^n)\to H_n(\mathbb{S}^n)$ is given by $H_n(f)[x]=d[x]$, where $[x]$ is a fixed generator for $H_n(\mathbb{S}^n)\cong \mathbb{Z}$, being your case $n=1$. Sorry for the mess.
Also, note that there is no contradiction in the theorem because $G$ is being restricted to $\mathbb{S}^1\times \{0\}$, which is homeomorphic to $\mathbb{S}^1$ and therefore it makes sense to talk about degree.
Best Answer
Embedding in $\mathbb C$, the identity is $z\mapsto z^1$ and the constant map is $z\mapsto 1=z^0$. If $(z\mapsto z^n)\simeq(z\mapsto z^m)$ let $\pi_t:S^1\times I\to S^1$ be such a homotopy with $$\pi_0=(z\mapsto z^n),\pi_1=(z\mapsto z^m).$$ Lift to a homotopy $\bar\pi_t$ of paths in $\mathbb R$ starting at $0$, noting $\bar\pi_0=\bar p_n$ and $\bar\pi_1=\bar p_m$ where $p_i$ is $z\mapsto z^i$. Then $\bar\pi_t(1)$ is independent of $t$, so $$n=\bar\pi_0(1) = \bar\pi_1(1)=m.$$
"Now use $0\neq 1$." This answer uses only elementary properties of covering spaces.
Note: in the last equation, the first and third equalities come from the fact that $n$th power mappings in $S^1$ lift to $s\mapsto ns$ in $\mathbb R$.