Adhesion is not stronger at faster peeling rate. If we formally associate adhesion as "the amount of energy needed to separate two parts", then we can describe it physically by the potential joining those two parts together.
In classical mechanics, where potentials are independent of time, like in gravitation or electrostatics, you can still explain why it "feels harder" to separate two parts the faster you do it: it "feels harder" because the force needed is bigger.
But why do we need bigger force to separate two parts attached by a potential which is independent of the external force applied or the time?
Because the effect expected (parts separated to a non-interaction distance) involves using an amount of energy equal to the potential energy bounding the parts. And although this amount of energy is independent of time, the force needed does depend on the time desired.
In mathematical expression, the work performed by the external force:
$$W_e = \int_0^d F_e(s)ds$$
is a time independent magnitude which does not care about the force involved nor its time dependence. But if we put $ds = v(t)dt$ we get:
$$W_e = \int_0^{t_d} F_e(t) v(t)dt$$
it can be seen how, the shorter is the time desired to achieve full separation $t_d$, the higher values of the integrand is needed. This integrand is a magnitude of power, or energy per unit of time, thus the faster the separation, the higher the power needed.
However, the power is the product of the force and velocity here $F_e(t) v(t)$, so is not clear yet why the force has to be higher. After all, a higher $v(t)$ would also contribute to a higher power so $F_e(t)$ does not have to be necessarily higher for faster separation.
Well actually the speed of separation depends on the balance of the external force to the force related to the bounding potential $F_b$:
$$v(t) = \int_0^{t_d} \frac{F_e - F_b}{m_a}dt$$
which is always smaller than the maximum velocity $v(t) < v_m(t)$ that can be achieved by $F_e(t)$, which is:
$$v_m(t) = \int_0^{t_d} \frac{F_e}{m_a}dt$$
and therefore, whatever the force you apply $F_e(t)$
$$F_e(t)v(t)<F_e(t)v_m(t)$$
it will not be as effective in increasing speed, as if there were no bounding potential (nor the related force $F_b(t)$).
Thus although $v(t)$ will be higher the faster is the removal, it will not suffice to account for the necessary power, and an increase in $F_e(t)$ will always occur such that the total work performed is constant however small the time employed.