Too long for a comment. Your conjecture is correct: $q\equiv-1\pmod p$ is a necessary and sufficient condition. This follows directly from the two lemmas below.

It's not a silly question, but it can be answered with some standard tools of representation theory, and a couple of facts about $Sp_{2r}(p)$: It is perfect and has trivial Schur multiplier. (I'll ignore the case $r=1$, $p=3$, for which $Sp_{2r}(p)\cong SL_2(3)$; subgroups of $GU_3(q)$ are well-known, thanks to H. H. Mitchell.) Let $q$ be a prime power which is relatively prime to the odd prime $p$. Let $r$ be a positive integer.  

Consider $p^{1+2r}$, which refers, I assume, to the extraspecial $p$-group of exponent $p$ and cardinality $p^{1+2r}$. Let $q$ be a power of the prime $s$, $s\ne p$. 

Lemma 1. $P:=p^{1+2r}$ embeds in $L:=GU_{p^r}(q)$ (via a monomorphism $\rho$, say) if and only if $q\equiv-1\pmod p$. 

Necessity: By the well-known characteristic $0$ representation theory of $P$, and the well-known connection between characteristic $0$ and characteristic $s$ representation theories of $P$ (since $s\ne p$), any faithful $p^r$-dimensional representation $\rho$ of $P$ in characteristic $s$ is absolutely irreducible.  Then $\rho(Z(P))$ consists of scalar matrices. The scalar subgroup of $GU_m(q)$ for any $q$ is homocyclic abelian of exponent $q+1$, so $p=|Z(P)|$ divides $q+1$. 

Sufficiency: Suppose that $p$ divides $q+1$. Let $P_0$ be an elementary abelian subgroup of $P$ of maximal order. Then $|P_0|=p^{r+1}$ and $Z(P)\le P_0$. Write $P_0=Z(P)\times P_1$ and let $\sigma:P_0\to GU_1(q)$ be a representation with kernel $P_1$. Such a representation exists since $p$ divides $q+1$. Induce $\sigma$ to $P$ to obtain an embedding $\rho:P\to GU(p^r,q)$. 

Lemma 2. Let $V$ be a $p^r$-dimensional vector space over $GF(q^2)$ equipped with a nondegenerate hermitian form. (There exists a unique such $V$, up to isometry.) Then any embedding $\rho:P\to GU(V)$ extends to an embedding $\sigma:P:Sp_{2r}(p)\to GU(V)$. 

To prove Lemma 2, first observe that there is an extension to $\sigma_0:P:Sp_{2r}(p)\to GL(V)\cong GL_{p^r}(q^2)$. 
To see this, observe that for every $g\in Sp_{2r}(p)$, the representations $\rho$ and $\rho^g$ are equivalent over $GL_{p^r}(q^2)$. Here $\rho^g(x)=\rho(g^{-1}xg)$. Since $g$ centralizes $Z(P)$, $\rho^g$ and $\rho$ have the same character, proving the equivalence; and both are absolutely irreducible. By definition of equivalence, there is an $X(g)\in GL(V)$ such that $\rho^g(x)=X(g^{-1})\rho(x)X(g)$ for all $x\in P$. By the absolute irreducibility and Schur's Lemma, $X(g)$ is uniquely determined by this condition, up to multiplication by a scalar. Then (making arbitrary choices for each $X(g)$) $X$ is a projective representation of $Sp_{2r}(p)$. Now the theory of projective representations, plus the facts that $Sp_{2r}(p)$ has trivial Schur multiplier, implies that the $X(g)$'s, $g\in Sp_{2r}(p)$, may be chosen so that $X$ is a genuine representation. Then define $$\sigma_0(xg)=\rho(x)X(g)$$ 
for every $x\in P$ and $g\in Sp_{2r}(p)$. Since $\rho$ and $X$ are representations, and $\rho(g^{-1}xg)=X(g)^{-1}\rho(x)X(g)$, $\sigma_0$ is a representation of the semidirect product. 

Finally, again by Schur's Lemma, $\rho(P)$ preserves a  unique hermitian form $H$ on $V$, up to multiplication by a scalar in $GF(q)$. This implies that for each $g\in Sp_{2r}(p)$, $X(g)$ carries $H$ to $\lambda(g)H$ for some $\lambda(g)\in GF(q)^\times$. Since $\sigma_0$ is a representation, so is $\lambda$. But $Sp_{2r}(p)$ is perfect, so $\lambda(g)=1$ for all $g\in Sp_{2r}(p)$. That is, $\sigma_0$ preserves $H$, i.e., $\sigma_0(P:Sp_{2r}(p))\le GU_{p^r}(q)$. And $\sigma_0$ obviously extends $\rho$, as required.