I'm interpreting your question as follows: when you write $H = \pi_1(Y,\ast)$, you mean the inclusion $i:Y\rightarrow X$ induces an injective map on $\pi_1$ with image $H$. (As opposed to $\pi_1(Y)$ being abstractly isomorphic to $H$).
With this interpretation, the answer is no. For example, consider $X = S^1$. It is well known that $\pi_1(X) = \mathbb{Z}$. Now, consider the subgroup $H = 2\mathbb{Z}$. I claim that there is no proper subspace $Y$ for which which has this fundamental group.
We may assume wlog that $Y$ is connected. Then note that any connected proper subset of $S^1$ is homeomorphic to a connected subset of $(0,1)$. These are easy to classify, they are, up to homeomorphism, $(0,1)$, $[0,1]$, and $(0,1]$. None of these has infinite cyclic fundamental group.
Edit Here is an example with $H$ not even abstractly isomorphic to a particular subgroup of $\pi_1(X)$.
Take $X = S^1 \vee S^1$, the wedge sum of 2 $S^1$s. The fundamental group of $X$ is known to be isomorphic to the free group on two generators. It's also know that a free group on two generators contains subgroups isomorphic to the free group on $n$ generators for any finite $n$ (we make even take $n$ to be countable). Further, it's know that free groups on different numbers of generators are never isomorphic.
Let $H$ denote any of these subgroups for $n > 2$. In particular, $H$ is not isomorphic to either $0$, $\mathbb{Z}$, or the free group on two generators. I claim that no subspace $Y$ of $X$ has $H$ as a fundamental group, even up to abstract isomorphism.
As above, we may assume $Y$ is connected. If $Y$ does not contain the wedge point, then it must be contained in a proper portion of one of the two circles, so the above argument shows $\pi_1(Y)$ is trivial. Hence, $Y$ must contain the wedge point. Now, if $Y$ does not contain the whole of the first circle, then $Y$ deformation retracts onto a subspace of the other circle, hence, by the previous argument has $\pi_1 = 0$ or $\mathbb{Z}$. Thus, $Y$ must contain the while first circle. Likewise, $Y$ must contain the whole second circle.
But then this implies $Y = X$, so $Y$s fundamental group is isomorphic to the free group on $2$ generators, so not isomorphic to $H$.
Final (?) Edit Here's one in the finite fundamental group case. Let $X$ be obtained from $S^1$ by attaching a $D^2$ by a degree $4$ map. A simple van Kampen argument shows $\pi_1(X) = \mathbb{Z}/4$. Let $H$ be the unique subgroup of $\pi_1(X)$ isomorphic to $\mathbb{Z}/2$.
I claim that no subspace $Y$ has fundamental group abstractly isomorphic to $H$. If $Y$ misses a point of the interior of the $D^2$, then $Y$ deformation retracts onto a subspace of $S^1$, so by the above argument, has $\pi_1 = 0$ or $\mathbb{Z}$. Hence, we may assume wlog that $Y$ contains all of the interior of $D^2$. Likewise, if $Y$ misses a point of $S^1$, then restricting the van Kampen argument to $Y$ shows that $\pi_1(Y) = 0$, so $Y$ must contain all of $S^1$. This implies $Y = X$, so $\pi_1(Y)\neq H$.
Your conclusion that if $\alpha, \beta : x_0 \rightarrow x_1$, then $\alpha \simeq_p \beta$(since $X$ is path-connected) is wrong. (I think you mean path homotopy). It is true that any two paths in path connected space is homotpic to each other but not path-homotopy.
$p:x_0 \rightarrow x_1$ and $q:y_0 \rightarrow y_1$ are any two paths in path-connected space $X$. Then $p \simeq e_{x_0}$ and $q \simeq e_{y_0} $ and since, $e_{x_0} \simeq e_{y_{0}} \Rightarrow p \simeq q $. But this doesn't mean that $p \simeq_p q$. There is a difference between path-homotopy and homotopy that path-homotopy has to satisfy additional criteria of initial point and final point for each $t \in I$, which clearly doesn't get satisfied in the proof of homotopy ($\textit{free homotopy}$).
Best Answer
Hint: there is $r: \mathbb{R}^2\setminus\{0\} \to S^1$ such that $S^1\to \mathbb{R}^2\setminus\{0\}\to S^1$ is the identity on $S^1$; and $S^1\to X\to \mathbb{R}^2\setminus \{0\} = S^1\to X\to \mathbb{R}^2\setminus\{0\}$