Consider the linear ordinary differential equation: \begin{align*} \dot x(t) & = Ax(t), \\ x(0) & = x_0, \end{align*} $t\geq 0$, $x\in\mathbb{R}^n$, with $A$ is Hurwitz (i.e., all its eigenvalues have negative real part).
Then, there exists a $\Gamma > 0$ and an $\eta > 0$ such that, \begin{align*} \Vert x(t) \Vert & \leq \Gamma e^{-\eta t}\Vert x_0 \Vert,\quad\forall t\geq 0,\quad \forall x_0\in\mathbb{R}^n \end{align*}
Consider now the ODE: \begin{align*} \dot y(t) & = f(y(t)) , \\ y(0) & = y_0, \end{align*} $t\geq 0$, $y\in\mathbb{R}^n$, with $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ a smooth function satisfying, \begin{align*} f(y) \leq Ay,\quad \forall y\in\mathbb{R}^n. \end{align*}
Is it true that there exists a $\Gamma_2 > 0$ and an $\eta_2 > 0$ such that: \begin{align*} \Vert y(t) \Vert & \leq \Gamma_2 e^{-\eta_2 t}\Vert y_0 \Vert,\quad\forall t\geq 0, \quad \forall y_0\in\mathbb{R}^n\quad? \end{align*}
My attempts:
I am trying to use the "comparison principle". This states that, if you have a scalar system $\dot u(t) = f(u(t))$, $u(0) = u_0$, and if you know that $\dot v(t) \leq f(v(t))$, $v(0) \leq u(0)$ for all $t\geq 0$, then $v(t) \leq u(t)$ for all $t\geq 0$.
Let $v := \frac{1}{2}\Vert x\Vert^2$. Then,
\begin{align*} \dot{v} & = x^\top \dot x = x^\top A x. \end{align*}
With, for example, $n = 2$, let,
\begin{equation} A = \left(\begin{array}{cc} a & b \\ c & d \end{array} \right) \end{equation} Then, \begin{align*} \dot{v} & =x_1(ax_1 + bx_2) + x_2(cx_1 + dx_2) \leq \max\{a,d\}(x_1^2 + x_2^2) + (b + c)x_1x_2. \end{align*}
Now, if $b = c = 0$ (which implies $a,d <0$ because $A$ is Hurwitz), then \begin{align*} \dot{v} & \leq 2\max\{a,d\}v, \end{align*}
and I can say, \begin{align*} v(t) & \leq e^{(2\max\{a,d\})t}v(0)\quad \forall v(0)\in\mathbb{R}^2. \end{align*}
I am stuck with when $b$ and $c$ are not zero... how do I deal with the cross terms?
Consider $\dot{x}=-x$ and $\dot{y}=f(y):=-y-y^2$. Those systems satisfy your conditions. However, we have that
$$x(t)=e^{-t}x(0)$$ while
$$y(t)=\dfrac{e^{-t}y(0)}{1+y(0)(1-e^{-t})}.$$
If we now choose $y(0)<-1$, then the second system has a finite escape time. This proves that the implication you want to show is not true in general.