Yes it is true.

Assume contrary, then there are two sequences of points $(x_n)$ and $(y_n)$ such that $$|p-x_n|=|p-y_n|=\tfrac1n$$
and
$$\measuredangle [p^{x_n}_{y_n}]-\tilde\measuredangle (p^{x_n}_{y_n})\ge \epsilon$$
for some fixed $\varepsilon>0$.
We can assume that the directions of $[px_n]$ and $[py_n]$ converge correspondingly to $\xi$ and $\upsilon$ in $\Sigma_p$.
Set
$$\alpha=\measuredangle(\xi,\upsilon)=\lim_{n\to\infty}\measuredangle [p^{x_n}_{y_n}]$$

Choose two geodesics $[pa]$ and $[pb]$ in the directions close to $\xi$ and $\upsilon$ correspondingly.
Let $a_n\in[pa]$ and $b\in[pb]$ be the points such that
$|p-a_n|=|p-b_n|=\tfrac1n$.

Note that for large $n$ we have
$$\lim_{n\to\infty}\tilde\measuredangle (p^{a_n}_{b_n})\approx \alpha.$$
By comparison, the values $n\cdot |a_n-x_n|$ and $n\cdot|y_n-b_n|$ are small.
It follows that
$$\tilde\measuredangle (p^{x_n}_{y_n})\approx\tilde\measuredangle (p^{a_n}_{b_n}).$$
Therefore
$$\lim_{n\to\infty}\tilde\measuredangle (p^{x_n}_{y_n})\approx\alpha,$$

a contradiction.