Let $C$ be a projective smooth curve over $\mathbb{Q}_p^\text{ur}$.
I wonder if it is possible that $C(\mathbb{Q}_p^\text{ur})=\emptyset$.
Here is my random guess:
Consider any regular model $\mathscr{C} \longrightarrow \operatorname{Spec}\mathbb{Z}_p^\text{ur}$ of $C \longrightarrow \operatorname{Spec}\mathbb{Q}_p^\text{ur}$. The special fibre is a curve over $\overline{\mathbb{F}}_p$ and it can not have a smooth point (otherwise Hensel's lemma would give a $\mathbb{Q}_p^\text{ur}$-point of the generic fibre).
I can imagine that for some particular model, every irreducible component of the special fibre has multiplicity $\geq 2$.
But I can not imagine that special fibre of every regular model has no smooth point.
For example, the curve $C: X^3 + pY^3 + p^2Z^3 = 0$ is smooth and doesn't have a $\mathbb{Q}_p^\text{ur}$-point.
If $[x, y, z]$ is a $\mathbb{Q}_p^\text{ur}$-point on $C$, then we may choose the projective coordinates in such a way that $\min(v(x), v(y), v(z)) = 0$, where $v$ stands for the $p$-adic valuation.
From the equality $x^3 + py^3 + p^2z^3 = 0$, we see that $v(x) > 0$, and hence we can write $x = px'$.
Substituing into the equation, we again find $v(y) > 0$ and hence $y = py'$; and then $v(z) > 0$.
This contradicts the assumption that $\min(v(x), v(y), v(z)) = 0$.