A real-orientable ring spectrum $F$ admits a ring map from $MO$, and there is a straightforward ring map $MU\to MO$, so $F$ is also complex orientable. Moreover, $MO$ is a wedge of $H/2$'s, so $MO\wedge F$ is also a wedge of $H/2$'s, and $F$ is a retract of $MO\wedge F$ (by using the ring structure etc) so it is again a wedge of $H/2$'s. Thus, you don't expect to learn anything about (non)immersion from $F$ that you could not already learn from $H/2$ (although some kinds of bookkeeping may be simplified).
The mapping spectrum $E=F(S^1_+,MU)$ is complex-orientable but not real-orientable and has $$\tilde{E}^1(\mathbb{R}P^\infty) = \tilde{MU}^1(\mathbb{R}P^\infty) \oplus \tilde{MU}^0(\mathbb{R}P^\infty)$$ Here $MU^{\ast}(\mathbb{R}P^\infty)=MU^*[[x]]/{[2]}(x)$ with $|x|=2$ and $MU^{-2}=MU_2\simeq\mathbb{Z}$ so the first summand is zero but the second is not.
A real-orientable ring spectrum $F$ admits a ring map from $MO$, and there is a straightforward ring map $MU\to MO$, so $F$ is also complex orientable. Moreover, $MO$ is a wedge of $H/2$'s, so $MO\wedge F$ is also a wedge of $H/2$'s, and $F$ is a retract of $MO\wedge F$ (by using the ring structure etc) so it is again a wedge of $H/2$'s. Thus, you don't expect to learn anything about (non)immersion from $F$ that you could not already learn from $H/2$ (although some kinds of bookkeeping may be simplified).