Inverse image of prime ideal in noncommutative ring

The answer to your question is no. Take a prime ring $R$ and $R'\subseteq R$ a subring which is not prime. Then the inverse image of $(0)$ is $(0)$ which is not a prime ideal of $R'$. An example is the following: $R=\begin{pmatrix} \mathbb Z & n\mathbb Z\\ \mathbb Z & \mathbb Z\end{pmatrix}$ and $R'=\begin{pmatrix} \mathbb Z & n\mathbb Z\\ 0 & \mathbb Z\end{pmatrix}$ with $n\ge 1$. (This example is taken from Lam, Exercises in Classical Ring Theory, Exercise 10.7.)

By looking at the induced maps on quotient rings, the question is equivalent to: Is every subring of a prime ring also prime? There are lots of counterexamples. For example, if $R$ is an integral domain, then $M_2(R)$ is a prime ring, but the subring of diagonal matrices $\cong R \times R$ is not.