Yes, this is true.
Let $a\in A_+$. By lower semicontinuity it suffices to show that $t_1((a-\delta)_+) = t_2((a-\delta)_+)$ for all $\delta>0$ (where $(a-\delta)_+$ is the positive part of $a-\delta 1$ in the unitisation of $A$ (which is an element of $A$ and not actually in the unitisation)). Fix $\delta>0$.
The key trick for getting the approximations to work, is to use that $t_i$ extends canonically to a positive linear functional on $\mathrm{span} R_i^\ast R_i$. Hence if $x\in R_i$ and $b,c\in A_+$ then
\begin{equation}
- \| b-c\| t_i(x^\ast x) \leq t_i(x^\ast(b-c)x) \leq \| b-c\| t_i(x^\ast x)
\end{equation}
and thus
\begin{equation}
|t_i(x^\ast b x) - t_i(x^\ast c x) | \leq \| b-c\| t_i(x^\ast x).
\end{equation}
This will be used a couple of times (in particular, using that $(a-\delta)_+^{1/2}\in R_i$, since $(a-\delta)_+$ is in the Pedersen ideal; the smallest dense two-sided ideal of $A$).
Let $\epsilon>0$. Pick $e\in A_+$ be such that $(a-\delta)_+ e = (a-\delta)_+$, and let $x\in B$ be close enough to $e^{1/2}$ that $\| e - x^\ast x\| \max\{ t_1((a-\delta)_+) , t_2((a-\delta)_+)\} < \epsilon$. By the trick mentioned above we have
\begin{equation}
t_i((a-\delta)_+) = t_i((a-\delta)_+^{1/2} e (a-\delta)_+^{1/2}) \approx_\epsilon t_i((a-\delta)_+^{1/2} x^\ast x (a-\delta)_+^{1/2}).
\end{equation}
Now pick $y\in B$ close enough to $(a-\delta)_+^{1/2}$ so that
\begin{equation}
\| y y^\ast - (a-\delta)_+ \| \max\{ t_1(x^\ast x) , t_2(x^\ast x)\} < \epsilon.
\end{equation}
Then by the above trick
\begin{equation}
t_i((a-\delta)_+^{1/2} x^\ast x (a-\delta)_+^{1/2}) = t_i(x(a-\delta)_+ x^\ast) \approx_\epsilon t_i (xyy^\ast x^\ast).
\end{equation}
As $xy \in B$ we get $t_1(xyy^\ast x^\ast) = t_2(xyy^\ast x^\ast)$ by assumption, and therefore
\begin{equation}
| t_1((a-\delta)_+) - t_2((a-\delta)_+) | < 4 \epsilon.
\end{equation}
As $\epsilon>0$ was arbitrary we get $t_1((a-\delta)_+) = t_2((a-\delta)_+)$, and thus $t_1(a) = t_2(a)$ by lower semicontinuity.
