Not without further assumptions: for a very simple counterexample, let $T$ be the empty theory in the empty language, $T_1$ the theory $\{\exists x\,R(x)\}$ in language $\{R(x)\}$, and $T_2=\{\neg\exists x\,R(x)\}$ in the same language.

However, if the language extensions are disjoint, then the result is true (and somewhat nontrivial). That is, if $T$ is a theory in a language $\Sigma$, and $T_1$ and $T_2$ are conservative extensions of $T$ in languages $\Sigma_1$ and $\Sigma_2$ (respectively) such that $\Sigma_1\cap\Sigma_2=\Sigma$, then $T_1\cup T_2$ is a conservative extension of $T$. This is a form of Robinson’s joint consistency theorem.

To derive it from a more common formulation of the joint consistency theorem that requires $T$ to be complete, let $\phi$ be any $\Sigma$-sentence such that $T\nvdash\phi$; we will show $T_1\cup T_2\nvdash\phi$. Since $T\nvdash\phi$, there exists a complete consistent $\Sigma$-theory $T'\supseteq T\cup\{\neg\phi\}$. Since $T_1$ and $T_2$ are conservative over $T$, the theories $T_1\cup T'$ and $T_2\cup T'$ are consistent. But then $T_1\cup T_2\cup T'$ is also consistent by the joint consistency theorem, hence $T_1\cup T_2\nvdash\phi$.

Yes, this is true (and somewhat nontrivial). That is, if $T$ is a theory in a language $\Sigma$, and $T_1$ and $T_2$ are conservative extensions of $T$ in languages $\Sigma_1$ and $\Sigma_2$ (respectively) such that $\Sigma_1\cap\Sigma_2=\Sigma$, then $T_1\cup T_2$ is a conservative extension of $T$. This is a form of Robinson’s joint consistency theorem.

To derive it from a more common formulation of the joint consistency theorem that requires $T$ to be complete, let $\phi$ be any $\Sigma$-sentence such that $T\nvdash\phi$; we will show $T_1\cup T_2\nvdash\phi$. Since $T\nvdash\phi$, there exists a complete consistent $\Sigma$-theory $T'\supseteq T\cup\{\neg\phi\}$. Since $T_1$ and $T_2$ are conservative over $T$, the theories $T_1\cup T'$ and $T_2\cup T'$ are consistent. But then $T_1\cup T_2\cup T'$ is also consistent by the joint consistency theorem, hence $T_1\cup T_2\nvdash\phi$.

The requirement $\Sigma_1\cap\Sigma_2=\Sigma$ is essential, otherwise there are very simple counterexamples: e.g., let $T$ be the empty theory in the empty language, $T_1$ the theory $\{\exists x\,R(x)\}$ in language $\{R(x)\}$, and $T_2=\{\neg\exists x\,R(x)\}$ in the same language.