It is consistent that the answer is no. The following is proved in Beaodouin's thesis ``On uncountable trees and linear orders'', as Theorem 1.10:
Theorem. Assume $\kappa^{<\kappa}=\kappa$ and $\Diamond(E)$ holds, where $E \subseteq \{\alpha < \kappa^+: cf(\alpha)=\kappa \}$.Then there is a normal $\kappa^+$-Aronszajn tree which has no special or Souslin subtree of size $\kappa^+.$
Theorem. Assume $\kappa^{<\kappa}=\kappa$ and $\Diamond(E)$ holds I may mention that the theorem is not stated as above in the paper, where $E \subseteq \{\alpha < \kappa^+: cf(\alpha)=\kappa \}$but looking at the proof, it is clear that the above result is in fact proved.Then there It is a normalalso mentioned in the paper that the case $\kappa^+$-Aronszajn tree which$\kappa=\omega$ has no special or Souslin subtree of size $\kappa^+.$discovered by Hanazawa too.