Analytic p-adic functions that take an algebraic value

Suppose it exists $$r\in\mathbb R$$ such that the non constant p-adic function $$f(z)=\sum_{n\ge0}a_nz^n$$ ($$a_n\in\mathbb C_p$$) is defined on $$\mathcal D=\{z\in\mathbb C_p\mid v_p(z)>r\}$$. Does it exist $$\alpha\in\overline{\mathbb Q}\cap f(\mathcal D)$$? If the answer is yes, does it exist $$\alpha\in{\overline{\mathbb Q}}\cap f(\mathcal D\cap\mathbb Q_p)$$?

For the first, yes. Without loss of generality by shifting we may assume $$a_1 \neq 0$$.
For $$\alpha\in \mathbb Q$$, let $$x_0=0$$ and $$x_{n+1} = x_n + \frac{\alpha-f(x)}{a_1}$$. To check that $$x_n$$ converges as $$n$$ goes to infinity to a root of $$\alpha-f(x)$$, it suffices to check that $$v_p ( \alpha - f(x_{n+1})) \geq v_p ( \alpha - f(x_n)) + 1$$.
To do this, it suffices to have $$v_p(x_n) \geq s$$ and $$v_p( x_{n+1}- x_n) \geq s$$ for some $$s \in \mathbb R$$ such that $$v_p (a_n)+n s> v_p (a_1) + s + 1$$ for all $$n > 1$$, as then the contributions of $$a_2$$ and higher to $$\alpha - f(x_{n+1})$$ will be dominated by the contribution of $$a_1$$.
This is easy to ensure by first choosing such an $$s < \frac{ v_p (a_n) - v_p(a_1) - 1}{ n-1}$$ for all $$n>1$$ (checking that this series is bounded below) and then choosing $$alpha$$ such that $$v_p ( \alpha- a_0) > a_1 + s$$, so that $$v_p(x_1-x_0)>s$$ and thus inductively $$v_p(x_{n+1} -x_n) >s$$ for all $$n$$.
For the second, no. Just take $$f(z) = z + b$$ where $$b\notin \mathbb Q_p + \overline{\mathbb Q}$$. Since $$\mathbb C_p/\mathbb Q_p$$ is uncountable, such $$b$$ exists.