Setting We work on a filtered probability space with finite time horizon $T$ and let $Q$ be a probability measure equivalent to $P$. Let $\rho_t=E[dQ/dP |\mathcal{F}_t]$, then $\rho$ is a martingale.
Result A process $M$ is a $Q$-martingale if and only if $\rho M$ is a $P$-martingale. See for example Lemma 5.3.4 in this book.
Question Does the same hold if we replace "martingale" by "submartingale", i.e. is $M$ a $Q$-submartingale if and only if $\rho M$ is a $P$-submartingale?
Proof (this is the proof of Lemma 5.3.4 in the above book adapted to the submartingale setting) Let $M$ be a $Q$-submartingale. Then $E[|\rho_t M_t|]=E[\rho_T |M_t|]=E_Q[|M_t|]<\infty$, showing that $\rho_t M_t$ is $P$-integrable. Next, for $s\leq t$ and $A\in\mathcal{F}_s$, $$E[\rho_tM_t1_A]=E[\rho_TM_t1_A]=E_Q[M_t1_A]\geq E_Q[M_s 1_A]=E[\rho_T M_s 1_A]=E[\rho_sM_s 1_A].$$ It is not clear if the next step is indeed true. Thus $E[\rho_tM_t|\mathcal{F}_s]\geq \rho_sM_s$, showing that $\rho M$ is a $P$-submartingale.
Proof (of the step) Let $N :=\rho M$. We know $E[(N_t-N_s)1_A]\geq 0, \forall A\in\mathcal{F}_s$. Define $D:=\{E[N_t|\mathcal{F}_s]-N_s=E[N_t-N_s|\mathcal{F}_s]<0\}\in\mathcal{F}_s$. Thus $$0\leq E[(N_t-N_s)1_D]=E[E[N_t-N_s|\mathcal{F}_s]1_D].$$ This implies $P[D]=0$ and thus $E[N_t|\mathcal{F}_s]\geq N_s$ a.s.
What you think is correct !
I update my previously wrong answer that now shows a failing attempt to construct a counterexample. As I always do I mixed up $P$- and $Q$-martingale. Hopefully not anymore.
With a $Q$-Brownian motion $W_t$ the proces $M_t=e^{W_t}$ is a $Q$-submartingale.
Let the Radon-Nikodym density of $Q$ w.r.t. $P$ be the $Q$-martingale $$ \frac{dP}{dQ}=e^{aW_t-a^2t/2}\,. $$ The inverse of this is the $P$-martingale $$ \rho_t=\frac{dQ}{dP}=e^{-aW_t+a^2t/2}=e^{-a(W_t+at)-a^2t/2} $$ where $W_t+at$ is a $P$-Brownian motion by Girsanov's theorem.
The process $$ \rho_t\,M_t=e^{(1-a)W_t+a^2t/2}=e^{(1-a)(W_t+at)-(1-a)\,at+a^2t/2} $$ is under $P$ the product of the $P$-martingale $X_t:=e^{(1-a)(W_t+at)-(1-a)^2t/2}$ and the deterministic function $$ Y_t:=e^{(1-a)^2t/2-(1-a)\,at+a^2t/2}=e^{\textstyle\frac{1-2a+a^2-a+a^2+a^2}{2}\scriptstyle t}=e^{\textstyle\frac{1-3a+3a^2}{2}\scriptstyle t}. $$ This function is strictly increasing in $t$ for every $a$. Therefore, $$ \mathbb E_P[\rho_tM_t|{\cal F}_s]=\mathbb E_P[X_tY_t|{\cal F}_s]=X_sY_t>X_sY_s=\rho_sM_s. $$ That is, $\rho_tM_t$ is a $P$-submartingale.